Q: A photoelectric material having work-function ϕ₀ is illuminated with light of wavelength λ (λ < hc/ϕ₀). The fastest photoelectron has a de Broglie wavelength λd. A change in wavelength of the incident light by Δλ results in a change Δλd in λd. Then the ratio Δλd/Δλ is proportional to:

λd²/λ². The ratio Δλd/Δλ is related to the de Broglie wavelength λd and the incident light wavelength λ, and can be derived from the equations governing the photoelectric effect and the de Broglie wavelength of the electron.

Q: In a photoelectric experiment, metals P, Q, and R emit photoelectrons with maximum kinetic energies EP, EQ, and ER, respectively, such that EP = 2EQ = 2ER. The same monochromatic light source is used for metals P and Q, while a different monochromatic light source is used for metal R. The

6. The energy of the photons incident on metal R is calculated using the photoelectric equation and the given kinetic energy relationships, resulting in a photon energy of 6 eV.

Q: A metallic surface is illuminated by light of a specific wavelength, and the stopping potential required to halt the emitted photoelectrons is measured to be 6.0 V. When the surface is exposed to light with a wavelength four times longer and intensity reduced to half, the stopping potentia

1.72 × 10⁻⁷ m, 1.20 eV. Using the photoelectric equation, the stopping potential and wavelength relationship are analyzed. The initial wavelength is calculated as 1.72 × 10⁻⁷ m, and the work function is determined to be 1.20 eV based on the energy difference.

Q: A proton starts from rest and is accelerated through a potential difference V, acquiring a de Broglie wavelength lambda. Through what potential difference must an alpha particle be accelerated from rest so that it acquires the same de Broglie wavelength lambda?

V/8 volt. De Broglie wavelength lambda = h / sqrt(2mqV). For equal wavelengths, mₚ * e * V = 4mₚ * 2e * V', giving V' = V/8.

Q: A metal plate carrying a uniform surface charge density of +2*epsilon₀ C/m² (in SI units) is illuminated with photons of wavelength 620 nm. The work function of the metal is 1.5 eV. Assuming photoelectrons are emitted perpendicular to the plate, what is the maximum distance from the plate

1/2 m. The photon energy is approximately 2 eV and work function is 1.5 eV, so KE_max = 0.5 eV = 8*10⁻²⁰ J. The electric field from the plate is E = sigma/(2*epsilon₀) = 1 V/m, so the electron is decelerated at a = eE/mₑ. Using KE = e*E*d, d = 0.5 eV / (e * 1 V/m) = 0.5 m = 1/2 m.

Q: A sodium street lamp rated at 200 W emits yellow light of wavelength 0.6 μm. Only 25% of the electrical power is converted into light. If the number of yellow photons emitted per second is approximately 1.5 * 10^x, find x.

20. With 25% efficiency the lamp radiates 50 W of light. Each yellow photon carries energy hc/lambda = 3.315*10⁻¹⁹ J, so the photon rate is 50 / 3.315*10⁻¹⁹ ~ 1.51*10²⁰, giving x = 20.

Question 1

The photoelectric effect provides evidence for the quantum behavior of light because:

Accepted Answer

photoelectrons are not emitted if the light's frequency is below a certain threshold value. The photoelectric effect demonstrates that light behaves as discrete packets of energy (photons). Electrons are only emitted if the light's frequency exceeds a threshold, proving that energy transfer is quantized.

Question 2

A metal surface is exposed to light with wavelengths of 248 nm and 310 nm. The maximum velocities of the emitted photoelectrons for these wavelengths are v₁ and v₂, respectively. Given that the ratio v₁: v₂ is 2: 1 and hc = 1240 eV·nm, estimate the work function of the metal.

Accepted Answer

3.7 eV. The correct answer is 3.7 eV because we can use the Einstein photoelectric effect equation to estimate the work function of the metal, given the ratio of the maximum velocities of the emitted photoelectrons and the wavelengths of the light.

Question 3

An electron in a higher energy level of a Li²⁺ ion possesses an angular momentum of 3h/2π. The electron's de Broglie wavelength in this energy state is expressed as pπa₀ (where a₀ represents the Bohr radius). Determine the value of p.

Accepted Answer

2. The angular momentum quantization condition for a hydrogen-like atom is L = nh/2π. Given L = 3h/2π, the principal quantum number n = 3. The de Broglie wavelength is inversely proportional to the momentum, and for n = 3, the wavelength is 2πa₀, making p = 2.

Question 4

In an experiment to calculate Planck's constant, light with varying wavelengths was shone on a metal surface, and the energy of the emitted photoelectrons was determined using stopping potentials. The following data was recorded for the wavelength (λ) of the light and the corresponding sto

Accepted Answer

6.4 × 10⁻³⁴. Using the photoelectric equation, h = eV₀λ/c, the data points are used to calculate Planck's constant. The average value obtained from the data is 6.4 × 10⁻³⁴ J·s.

Question 5

A photoelectric material having work-function ϕ₀ is illuminated with light of wavelength λ (λ < hc/ϕ₀). The fastest photoelectron has a de Broglie wavelength λd. A change in wavelength of the incident light by Δλ results in a change Δλd in λd. Then the ratio Δλd/Δλ is proportional to:

Accepted Answer

λd²/λ². The ratio Δλd/Δλ is related to the de Broglie wavelength λd and the incident light wavelength λ, and can be derived from the equations governing the photoelectric effect and the de Broglie wavelength of the electron.

Question 6

In a photoelectric experiment, metals P, Q, and R emit photoelectrons with maximum kinetic energies EP, EQ, and ER, respectively, such that EP = 2EQ = 2ER. The same monochromatic light source is used for metals P and Q, while a different monochromatic light source is used for metal R. The

Accepted Answer

6. The energy of the photons incident on metal R is calculated using the photoelectric equation and the given kinetic energy relationships, resulting in a photon energy of 6 eV.

Question 7

A metallic surface is illuminated by light of a specific wavelength, and the stopping potential required to halt the emitted photoelectrons is measured to be 6.0 V. When the surface is exposed to light with a wavelength four times longer and intensity reduced to half, the stopping potentia

Accepted Answer

1.72 × 10⁻⁷ m, 1.20 eV. Using the photoelectric equation, the stopping potential and wavelength relationship are analyzed. The initial wavelength is calculated as 1.72 × 10⁻⁷ m, and the work function is determined to be 1.20 eV based on the energy difference.

Question 8

A proton starts from rest and is accelerated through a potential difference V, acquiring a de Broglie wavelength lambda. Through what potential difference must an alpha particle be accelerated from rest so that it acquires the same de Broglie wavelength lambda?

Accepted Answer

V/8 volt. De Broglie wavelength lambda = h / sqrt(2mqV). For equal wavelengths, mₚ * e * V = 4mₚ * 2e * V', giving V' = V/8.

Question 9

A metal plate carrying a uniform surface charge density of +2*epsilon₀ C/m² (in SI units) is illuminated with photons of wavelength 620 nm. The work function of the metal is 1.5 eV. Assuming photoelectrons are emitted perpendicular to the plate, what is the maximum distance from the plate

Accepted Answer

1/2 m. The photon energy is approximately 2 eV and work function is 1.5 eV, so KE_max = 0.5 eV = 8*10⁻²⁰ J. The electric field from the plate is E = sigma/(2*epsilon₀) = 1 V/m, so the electron is decelerated at a = eE/mₑ. Using KE = e*E*d, d = 0.5 eV / (e * 1 V/m) = 0.5 m = 1/2 m.

Question 10

A sodium street lamp rated at 200 W emits yellow light of wavelength 0.6 μm. Only 25% of the electrical power is converted into light. If the number of yellow photons emitted per second is approximately 1.5 * 10^x, find x.

Accepted Answer

20. With 25% efficiency the lamp radiates 50 W of light. Each yellow photon carries energy hc/lambda = 3.315*10⁻¹⁹ J, so the photon rate is 50 / 3.315*10⁻¹⁹ ~ 1.51*10²⁰, giving x = 20.

Question 11

Two particles, each having the same de Broglie wavelength lambda, collide and merge to form a single new particle. The new particle also has de Broglie wavelength lambda. What must be the angle between the initial directions of motion of the two particles?

Accepted Answer

120 deg. Conservation of momentum: |p_total|² = p1² + p2² + 2p1*p2*cos(theta) = p² + p² + 2p²*cos(theta). For the resultant momentum magnitude to be p (wavelength = lambda), we need p² = 2p²(1 + cos(theta)), giving cos(theta) = -1/2, so theta = 120 deg.

Question 12

A light source emitting only 400 nm wavelength (selected by an interference filter) shines on a metal surface and ejects electrons. The filter is then swapped for one that passes only 300 nm, and the lamp intensity falling on the metal is kept identical to before. Which statement correctly

Accepted Answer

The emitted electrons have greater kinetic energy.. At the same intensity, switching from 400 nm to 300 nm means each photon carries more energy (E = hc/lambda is larger), so there are fewer photons per second hitting the surface, meaning fewer photoelectrons are emitted. However, each emitted electron gains more kinetic energy (KE = hf - phi), so the electrons are more energetic.

Question 13

A laser device of power 3 mW and mass 50 g emits photons of wavelength 500 nm while floating in gravity-free space. It fires continuously for exactly one hour. What is the approximate distance the device travels due to recoil during this time?

Accepted Answer

1.3 mm. The laser imparts a recoil force F = P/c on the device (by Newton's third law). With constant force on mass m in free space, the acceleration a = P/(m*c) is constant, and the device travels distance s = (1/2)*a*t² in time t = 1 hour.

Question 14

A metallic sphere of radius R = 18 cm is illuminated with light of wavelength lambda = 4000 angstrom. The threshold wavelength for the sphere's material is lambda₀ = 6000 angstrom. A capacitor of capacitance C₀ = 2 microfarad is connected to the sphere by closing a switch. After a long tim

Accepted Answer

2 microcoulombs. The stopping potential is Vₛ = (hc/e)(1/lambda - 1/lambda₀) = 12*10⁻⁷ * (1/(4*10⁻⁷) - 1/(6*10⁻⁷)) = 12*10⁻⁷ * (2.5*10⁶ - 1.667*10⁶) = 12*10⁻⁷ * (8.33*10⁵) = 1 V. The sphere's own capacitance is C_sphere = 4*pi*epsilon₀*R = R/k = 0.18/(9*10⁹) = 2*10⁻¹¹ F = 0.00002 microfarad, which is negligible compared to C₀ = 2 microfarad. After a long time, the capacitor and sphere reach equilibrium potential Vₛ = 1 V. Charge on capacitor = C₀ * Vₛ = 2 * 10⁻⁶ * 1 = 2 * 10⁻⁶ C = 2 microcoulombs.

Question 15

An electron starts from rest and is accelerated through a potential difference V1, acquiring a de Broglie wavelength of 1.41 angstroms. If the potential difference is reduced so that the new wavelength becomes 1.73 angstroms, by how much has the potential difference decreased?

Accepted Answer

20 V. Since lambda * sqrt(V) = constant: V1/V2 = (lambda2/lambda1)² = (1.73/1.41)² approx (sqrt(3)/sqrt(2))² = 3/2. So V1 = (3/2)*V2. The decrease = V1 - V2 = V2/2. From V2 = 2*V1/3 and lambda1 = h/sqrt(2meV1), a specific V1 is needed. Using 1.41 A => V1 = 60 V, 1.73 A => V2 = 40 V, decrease = 20 V.

Question 16

A photon of wavelength 124 nm is absorbed by a metal whose work function is 6 eV. The maximum uncertainty in the momentum of the ejected photoelectron is (h/(4*pi)) * 10⁹ kg*m/s. Given that (6.135)² / pi = 12, identify the correct statements.

Accepted Answer

Maximum kinetic energy of the ejected photoelectron is 4 eV.. Photon energy is 10 eV, work function is 6 eV, so KE_max = 4 eV. Using Heisenberg's principle with the given deltaₚ, deltaₓ = 1 nm. The uncertainty in de-Broglie wavelength can be computed from delta_lambda/lambda = deltaₚ/p.

Question 17

An electron is confined within a tube of length L. The potential energy of the electron is zero in the first half of the tube and 12 eV in the second half. The total energy of the electron is 16 eV. Find the ratio of the longest to the shortest de Broglie wavelength of the electron inside

Accepted Answer

2. Since lambda is proportional to 1/sqrt(KE), the ratio of longest (lower KE) to shortest (higher KE) wavelength is sqrt(16)/sqrt(4) = 4/2 = 2.

Question 18

A photoelectron is emitted normally (perpendicularly) from the positive plate of a parallel plate capacitor. The potential difference across the capacitor is 1 V. The wavelength of incident light is 310 nm and the work function of the plate material is 2.2 eV. Given: mass of electron mₑ =

Accepted Answer

1 mm. Initial KE = hc/lambda - phi = 4 - 2.2 = 1.8 eV. Initial velocity u = sqrt(2*1.8*1.6e-19 / 9e-31). The electric field between plates accelerates the electron from positive to negative plate (force on electron is toward negative plate, same as its initial velocity direction). Using d = u*t + (1/2)*a*t² with t=3 ns gives d approximately 1 mm.

Question 19

The kinetic energy (KE) of a photoelectron emitted when light of frequency v strikes a metal surface obeys KE = hv - phi, where phi is the work function. Which of the following correctly describe(s) the graph of KE versus incident frequency?

Accepted Answer

A straight line whose slope equals Planck's constant h. The equation KE = hv - phi is linear in v. The slope is h (Planck's constant), the x-intercept occurs at the threshold frequency v₀ = phi/h (not h*v₀), and the extrapolated y-intercept is -phi (negative of threshold energy, not equal to it). Option A and D are correct; option B is incorrect (x-intercept = v₀, not h*v₀); option C states intercept equals threshold energy but it is actually -phi (negative).

Question 20

A photon of energy 8 eV strikes a metal surface whose threshold frequency is 1.6 * 10¹⁵ Hz. Find the maximum kinetic energy of emitted photoelectrons. (h = 6.6 * 10⁻³⁴ J s, 1 eV = 1.6 * 10⁻¹⁹ J)

Accepted Answer

1.4 eV. Work function phi = h * nu_threshold = 6.6e-34 * 1.6e15 = 10.56e-19 J = 10.56e-19/1.6e-19 eV = 6.6 eV. KE_max = E_photon - phi = 8 eV - 6.6 eV = 1.4 eV.

Question 21

A particle of mass m and charge q is accelerated through a potential difference of 50 V, producing a de Broglie wavelength lambda. If the mass is doubled (to 2m) while keeping the charge the same, and the new accelerating potential difference is 2500 V, what is the new de Broglie wavelengt

Accepted Answer

0.1 lambda. de Broglie wavelength: lambda = h/sqrt(2mqV). Original: lambda = h/sqrt(2*m*q*50). New: m' = 2m, V' = 2500 V. lambda' = h/sqrt(2*2m*q*2500) = h/sqrt(10000*m*q). Original denominator: h/sqrt(100*m*q). Ratio: lambda'/lambda = sqrt(100*m*q)/sqrt(10000*m*q) = sqrt(100/10000) = 1/10. So lambda' = 0.1*lambda.

Question 22

For photoelectric emission from a certain metal, the threshold frequency is n. When radiation of frequency 2n strikes the metal surface, what is the maximum speed of the emitted photoelectrons? (m = mass of electron)

Accepted Answer

sqrt(2 * h * n / m). Using Einstein's photoelectric equation: h * nu = phi + (1/2) * m * v². Here phi = h * n and nu = 2n, so (1/2) * m * v² = h * (2n) - h * n = h * n. Therefore v = sqrt(2 * h * n / m).

Question 23

In a photoelectric experiment, light of wavelength lambda1 = 3100 angstrom is used in the first setup and lambda2 = 6200 angstrom in the second setup, both on the same metal X. The ratio of maximum speeds of ejected electrons in the two setups is sqrt(5): 1. Find the work function of metal

Accepted Answer

1.5 eV. Energy of photon: E = hc/lambda. E1 = hc/3100A, E2 = hc/6200A = E1/2. Let phi = work function. KE1 = E1 - phi, KE2 = E2 - phi = E1/2 - phi. Given v1/v2 = sqrt(5), so KE1/KE2 = v1²/v2² = 5. So (E1 - phi)/(E1/2 - phi) = 5. Solve: E1 - phi = 5*(E1/2 - phi) = 5*E1/2 - 5*phi. 4*phi = 5*E1/2 - E1 = 3*E1/2. phi = 3*E1/8. E1 = hc/3100A = (6.626e-34*3e8)/(3100e-10) = (1.988e-25)/(3.1e-7) = 6.41e-19 J = 4.0 eV. phi = 3*4.0/8 = 1.5 eV.

Question 24

Which of the following statements concerning X-rays is/are correct?

Accepted Answer

The minimum cut-off wavelength depends solely on the accelerating voltage applied between the filament and the target. Option D is correct: lambda_min = hc/(eV), depends only on accelerating voltage V. Option A is incorrect because the energy-level relation E(K-alpha) + E(L-beta) is not generally equal across different series in that way. Option B is incorrect: harder X-rays have higher energy/shorter wavelength but the intensity depends on tube current, not on 'hardness'. Option C is incorrect: continuous (bremsstrahlung) and characteristic X-rays of the same wavelength are physically differe

Question 25

A photocell circuit has its cathode illuminated by monochromatic light. If the intensity of light is kept constant and only the frequency of the incident light is increased, which of the following statements is correct?

Accepted Answer

The maximum kinetic energy of the emitted photoelectrons increases. According to Einstein's photoelectric equation, maximum kinetic energy KE_max = hf - phi (work function). Increasing frequency f at constant intensity increases KE_max directly. The number of photons per second decreases (since each photon is more energetic), so photocurrent actually decreases slightly, but the definitive and always-true statement is that KE_max increases. Option (c) is unambiguously correct.

Question 26

An electron moves at a velocity of 1.5 * 10⁸ m/s. Its de Broglie wavelength equals the wavelength of a photon. What is the ratio of the kinetic energy of the electron to the energy of the photon?

Accepted Answer

1/4. Same wavelength: lambda. For photon: E = hc/lambda. For electron: lambda = h/(mv), so E_photon = hc/(h/mv) = mvc. KE_electron = (1/2)mv² (non-relativistic approximation). Ratio = (1/2)mv² / mvc = v/(2c) = (1.5e8)/(2*3e8) = 1.5/6 = 1/4.

Question 27

A solar sail spacecraft consists of a boat of mass m fitted with a large flat canvas of area A. The boat is accelerated by the radiation force from sunlight. Find the acceleration of the boat at a distance r from the Sun (power of the Sun = P; assume the sunlight strikes the canvas perpend

Accepted Answer

PA/(2*pi*r²*c*m). The solar intensity at distance r is I = P/(4*pi*r²). For a perfect reflector, radiation pressure = 2I/c. Force on canvas = 2IA/c = 2PA/(4*pi*r²*c) = PA/(2*pi*r²*c). Acceleration a = F/m = PA/(2*pi*r²*c*m).

Question 28

A monochromatic light source operating at a power of 200 W emits 4 * 10²⁰ photons per second. What is the wavelength of the light?

Accepted Answer

400 nm. Each photon has energy E = hc/lambda. Power P = N * E = N*h*c/lambda. So lambda = N*h*c/P = (4e20 * 6.626e-34 * 3e8) / 200 = (4e20 * 1.9878e-25) / 200 = (7.95e-5) / 200 = 3.975e-7 m ~ 400 nm.

Question 29

In a photoelectric experiment combined with a double-slit setup, the separation between slits is d = 0.24 mm, slit-to-screen distance D = 1.2 m, and the point of observation is y = 1.00 mm above the central maximum. The incident light has wavelength equal to the path difference at that poi

Accepted Answer

4.0 V. Path difference at y: Delta = yd/D = (1.00e-3)(0.24e-3)/1.2 = 2.0e-7 m = 200 nm. This is identified as the wavelength of the incident UV light. Photon energy = hc/lambda = 1240 eV*nm / 200 nm = 6.2 eV. By Einstein's equation: eV_stop = hf - phi = 6.2 - 2.2 = 4.0 eV. So stopping potential = 4.0 V.

Question 30

Particle A has mass m and charge q, accelerated through potential difference 50 V. Particle B has mass 4m and charge q, accelerated through potential difference 2500 V. Find the ratio of their de-Broglie wavelengths lambda_A / lambda_B.

Accepted Answer

14.14. lambda = h/sqrt(2mqV). Ratio = sqrt((m_B*V_B)/(m_A*V_A)) since q is the same. = sqrt((4m * 2500)/(m * 50)) = sqrt(10000/50) = sqrt(200) = 10*sqrt(2) ≈ 14.14.

Question 31

A metal plate is illuminated by light of wavelength lambda. Electrons are ejected from the plate surface. A uniform retarding electric field E is applied, and no electron can travel more than a distance d from the plate. If e is the electronic charge, h is Planck's constant, and c is the s

Accepted Answer

lambda0 = (1/lambda - e*E*d/(hc))^(-1). By Einstein's photoelectric equation: KE_max = hc/lambda - hc/lambda0 (where hc/lambda0 is the work function). The retarding field E stops the fastest electron within distance d, so KE_max = e*E*d. Therefore: e*E*d = hc/lambda - hc/lambda0. Rearranging: hc/lambda0 = hc/lambda - e*E*d. Dividing by hc: 1/lambda0 = 1/lambda - e*E*d/(hc). So lambda0 = 1/(1/lambda - e*E*d/(hc)) = (1/lambda - e*E*d/(hc))^(-1).

Question 32

A proton (p), a deuteron (d), and an alpha particle (alpha) each have the same de Broglie wavelength. What is the correct order of their kinetic energies?

Accepted Answer

E_alpha < E_d < Eₚ. Since lambda = h/sqrt(2*m*KE), for the same lambda: KE = h²/(2*m*lambda²). KE is inversely proportional to m. Masses: proton ~ 1u, deuteron ~ 2u, alpha ~ 4u. So KEₚ > KE_d > KE_alpha, i.e. E_alpha < E_d < Eₚ.

Question 33

How many photons per second of radiation with wavelength lambda = 5 * 10⁻⁷ m must strike a perfectly absorbing blackened plate to produce a force of 6.62 * 10⁻⁵ N? (h = 6.62 * 10⁻³⁴ J s)

Accepted Answer

5 * 10²². n = F*lambda/h = (6.62*10⁻⁵ * 5*10⁻⁷) / (6.62*10⁻³⁴) = 3.31*10⁻¹¹ / 6.62*10⁻³⁴ = 5*10²² photons/s.

Question 34

In a photoelectric effect experiment, the incident wavelength is lambda. When the wavelength is reduced by (2/3)*lambda (i.e., the new wavelength is lambda/3), the maximum kinetic energy of the emitted photoelectrons becomes n times the original value. What is the threshold wavelength of t

Accepted Answer

((n-1)/(n-3))*lambda. Let K = initial max KE, lambda0 = threshold wavelength. K = hc/lambda - hc/lambda0... (1). n*K = 3*hc/lambda - hc/lambda0... (2). Subtract (1) from (2): (n-1)*K = 2*hc/lambda => K = 2*hc/[(n-1)*lambda]. From (1): hc/lambda0 = hc/lambda - K = hc/lambda - 2*hc/[(n-1)*lambda] = hc/lambda * [1 - 2/(n-1)] = hc/lambda * (n-3)/(n-1). So lambda0 = lambda*(n-1)/(n-3).

Question 35

An electron moving with speed v and a photon moving with speed c have the same de Broglie wavelength. Let the kinetic energy and momentum of the electron be Eₑ and Pₑ, and those of the photon be Eₚ and Pₚ. Which statement is correct?

Accepted Answer

Eₚ / Eₑ = 2c / v. Same wavelength => same momentum: Pₑ = Pₚ = p. Electron kinetic energy (non-relativistic): Eₑ = p²/(2mₑ). Since p = mₑ*v (for non-relativistic electron), Eₑ = mₑ*v²/2. Photon energy: Eₚ = p*c. Ratio Eₚ/Eₑ = p*c / (p²/(2mₑ)) = 2mₑ*c/p = 2mₑ*c/(mₑ*v) = 2c/v. Since Pₑ = Pₚ, any ratio of momenta = 1, so options B and D are wrong. The correct statement is Eₚ/Eₑ = 2c/v.

Question 36

A fictitious hydrogen-like atom called 'Quantonium' has the same energy level formula as hydrogen (Eₙ = -13.6/n² eV). Photons emitted from transitions n=4 to n=2 and from n=2 to n=1 can eject photoelectrons from an unknown metal, but the photon emitted from n=3 to n=2 does not. What are th

Accepted Answer

5 eV < phi < 8 eV. Energy levels: Eₙ = -13.6/n² eV. E₁ = -13.6 eV, E₂ = -3.4 eV, E₃ = -1.51 eV, E₄ = -0.85 eV. Photon energies: n=4->2: E = E₂ - E₄ = -3.4 - (-0.85) =... wait, photon energy = E_higher - E_lower in magnitude = |E₄ - E₂| = |-0.85 - (-3.4)| = 2.55 eV. n=3->2: |E₃ - E₂| = |-1.51 - (-3.4)| = 1.89 eV. n=2->1: |E₂ - E₁| = |-3.4 - (-13.6)| = 10.2 eV. Conditions: n=4->2 (2.55 eV) ejects electrons: phi < 2.55 eV. n=2->1 (10.2 eV) ejects electrons: phi < 10.2 eV. n=3->2 (1.89 eV) does NOT eject: phi > 1.89 eV. So 1.89 < phi < 2.55. None of the options match exactly. Re-examining: ALLENAL

Question 37

Light of wavelength lambda strikes a photosensitive surface and electrons are ejected with kinetic energy E. To double the kinetic energy to 2E, the wavelength must be changed to lambda' where:

Accepted Answer

lambda/2 < lambda' < lambda. From photoelectric effect: E = hc/lambda - phi...(1). 2E = hc/lambda' - phi...(2). Subtracting (1) from (2): E = hc/lambda' - hc/lambda => hc/lambda' = hc/lambda + E = hc/lambda + (hc/lambda - phi) = 2hc/lambda - phi. So 1/lambda' = (2/lambda) - phi/(hc). Since phi > 0, 1/lambda' < 2/lambda, meaning lambda' > lambda/2. Also, since E > 0 means hc/lambda > phi > 0, so hc/lambda' = hc/lambda + E > hc/lambda, meaning lambda' < lambda. Therefore lambda/2 < lambda' < lambda.

Question 38

In a photoelectric effect experiment, a point source of light illuminates a metal surface. If the source is moved farther away from the metal, what happens to the stopping potential?

Accepted Answer

It remains constant.. The stopping potential equals eVₛ = h*f - phi, where f is the frequency of incident light and phi is the work function. Neither f nor phi changes when the source is moved farther away — only the intensity (rate of emission of photoelectrons) decreases. Hence the stopping potential remains constant.

Question 39

How many photons of yellow light (wavelength = 575 nm) must be absorbed by 1.0 cm² of skin to raise the skin temperature by 1.0 degree C? Express the answer as 1.2 * 10ⁿ and find n. (Treat tissue as water: density ~1 g/cm³, specific heat ~4.2 J/(g deg C); assume 1 cm³ of tissue.)

Accepted Answer

3. The heat needed to raise 1 cm³ of water by 1 deg C is Q = 1 g * 4.2 J/(g.K) * 1 = 4.2 J. Each photon carries energy E = hc/lambda = (6.626e-34 * 3e8)/(575e-9) ≈ 3.46e-19 J. Number N = 4.2 / 3.46e-19 ≈ 1.21e19. So n ≈ 19, but the question gives answer as 1.2 * 10ⁿ with options 1-4; re-reading options the question asks for n where the number is 1.2*10ⁿ and gives options 1,2,3,4 meaning the exponent digit. Given Q=4.2 J and E_photon~3.46e-19 J, N~1.2e19, so the exponent is 19, and if n is the units digit of the exponent (19 -> 9) or the question intends a simpler scenario. Checking: the option

Question 40

A hypothetical particle 'Xeton' at rest has a mass equal to that of a helium nucleus (4 proton masses). It absorbs n photons each of frequency nu₀. Its de Broglie wavelength is later measured to be (1/8) * sqrt(h / (m * nu₀)), where m is the mass of a proton. Find the value of n.

Accepted Answer

4. Momentum of each photon = h*nu₀/c. After absorbing n photons all in same direction, total momentum = n*h*nu₀/c. de Broglie wavelength lambda = h/(n*h*nu₀/c) = c/(n*nu₀). Given lambda = (1/8)*sqrt(h/(m*nu₀)). So c/(n*nu₀) = (1/8)*sqrt(h/(m*nu₀)). n = 8c/nu₀ * sqrt(m*nu₀/h) = 8c*sqrt(m/(h*nu₀)). This depends on values — the question requires relativistic treatment or specific mass ratio. Using mass of Xeton M = 4m and de Broglie for particle with momentum p: lambda = h/p. n = h/(lambda*h*nu₀/c) = c/(lambda*nu₀) = c / ((1/8)*sqrt(h/(m*nu₀)) * nu₀) = 8c / (nu₀ * sqrt(h/(m*nu₀))) = 8c * sqrt(m*n

Question 41

A collection of excited hydrogen-like single-electron ions can emit a maximum of 10 different spectral lines when de-exciting to the ground state. The maximum energy transition releases 117.5 eV. Photons corresponding to the minimum energy transition from this system strike a metal with wo

Accepted Answer

12.27. n*(n-1)/2 = 10 => n = 5. E_max = Z² * 13.6 * (1 - 1/25) = Z² * 13.6 * 24/25 = 117.5 eV => Z² * 13.056 = 117.5 => Z² = 9 => Z = 3 (Li²+ ion). Min energy transition: n=5 to n=4: E_min = 9 * 13.6 * (1/16 - 1/25) = 9 * 13.6 * 9/400 = 2.754 eV. KE of photoelectron = 2.754 - 1.254 = 1.5 eV. de Broglie: lambda = h/sqrt(2*m*KE). Using lambda = 12.27/sqrt(KE in eV) Angstrom = 12.27/sqrt(1.5) ~ 12.27/1.225 ~ 10.02 Angstrom. Actually standard formula: lambda (Angstrom) = sqrt(150/V_eV) = sqrt(150/1.5) = sqrt(100) = 10 Angstrom. Hmm. Let me re-check min energy: from n=5 to n=4 for Li²+: delta_E = 9

Question 42

Find the orbit number n of the electron in He⁺ whose de Broglie wavelength equals that of the electron in the 4th orbit of the hydrogen atom.

Accepted Answer

4. The de Broglie wavelength in orbit n of a hydrogen-like atom (atomic number Z) is lambda = 2*pi*n*a₀/Z. For hydrogen (Z=1, n=4): lambda_H = 2*pi*4*a₀ = 8*pi*a₀. For He⁺ (Z=2): lambda_He = 2*pi*n*a₀/2 = pi*n*a₀. Setting lambda_He = lambda_H: pi*n*a₀ = 8*pi*a₀, n = 8. Since 8 is not in the options (1,2,3,4), this question appears to have an error in the option set; the correct answer n=8 is not listed.

Question 43

A cricket ball of mass m is struck by a batsman and flies into the air; a fielder catches it near the boundary. The de Broglie wavelength of the ball just after being struck is lambda1 and just before being caught is lambda2. The ball is hit and caught at the same height. Find the total wo

Accepted Answer

h² / (2m) * (1/lambda1² - 1/lambda2²). De Broglie relation: lambda = h/p = h/(mv), so p = h/lambda and KE = p²/(2m) = h²/(2m*lambda²). Since initial and final heights are the same, delta(PE) = 0. By work-energy theorem, W_air = KE_final - KE_initial = h²/(2m) * (1/lambda2² - 1/lambda1²). Since air resistance slows the ball, lambda2 > lambda1 meaning 1/lambda2² < 1/lambda1², so W_air is negative (energy lost). Total work done BY air resistance = h²/(2m) * (1/lambda2² - 1/lambda1²) = -h²/(2m) * (1/lambda1² - 1/lambda2²). The option asking for total work done by air resistance (which is negative)

Question 44

In two Bohr orbits of a hydrogen-like species, the de Broglie wavelengths of the electron are in the ratio 3: 5. If the ratio of kinetic energies of the electron in these two orbits is 25: x, find the value of x.

Accepted Answer

9. In Bohr model, angular momentum mvr = n * h / (2 * pi) and r = n² * a0 / Z. So mv = n * h / (2 * pi * r) = Z * h / (2 * pi * n * a0). Thus lambda = h / (mv) = 2 * pi * n * a0 / Z, meaning lambda is proportional to n. Given lambda₁ / lambda₂ = 3 / 5, we get n1 / n2 = 3 / 5. Kinetic energy KE = m * e⁴ * Z² / (8 * epsilon0² * h² * n²) is proportional to 1 / n². So KE₁ / KE₂ = n2² / n1² = 25 / 9. Therefore x = 9.

Question 45

Two electrons move with the same speed v. One enters a region of uniform electric field, the other enters a region of uniform magnetic field. After some time, if their de Broglie wavelengths are lambda1 and lambda2 respectively, which option(s) are possible?

Accepted Answer

(B) lambda1 > lambda2. For the electron in the magnetic field: the magnetic force is always perpendicular to the velocity, so it does no work. The speed (and hence momentum) of the electron remains v. Therefore lambda2 = h/(mv) = initial wavelength. For the electron in the electric field: the electric field does work on the electron, changing its kinetic energy and hence its speed. Depending on the field direction relative to initial velocity, the electron can speed up or slow down. If it speeds up, momentum increases, lambda1 < lambda2. If it slows down, momentum decreases, lambda1 > lambda2.

Question 46

A photon of frequency v_i collides head-on with an electron of mass m moving with speed u_i. The photon is scattered in a direction exactly opposite to its initial direction of travel. It is observed that the frequency of the scattered photon equals the frequency of the incident photon. Wh

Accepted Answer

The magnitude of the initial momentum of the electron is p_i = h*v_i / c.. Let the photon travel in the +x direction. Initial photon momentum = +h*v_i/c. Final photon momentum = -h*v_i/c (reversed). Momentum conservation: p_i^e + h*v_i/c = p_f^e - h*v_i/c => p_f^e = p_i^e + 2h*v_i/c. Energy conservation (same photon frequency): E_i^e = E_f^e. Since the electron energy is unchanged but its momentum changed, we need (p_f^e)² = (p_i^e)² (from E²=(pc)²+(mc²)² with same E). So p_f^e = -p_i^e. Then: -p_i^e = p_i^e + 2h*v_i/c => p_i^e = -h*v_i/c. Magnitude: |p_i^e| = h*v_i/c. (A) correct. KE = p²/(2m

Question 47

A metal has threshold frequency f0. When light of frequency 2*f0 is incident on the metal, the maximum velocity of emitted photoelectrons is v1. When light of frequency 5*f0 is incident, the maximum velocity of emitted photoelectrons is v2. Find the ratio v1: v2.

Accepted Answer

1: 2. Einstein photoelectric equation: (1/2)m*v² = h*(f - f0). For f = 2f0: (1/2)m*v1² = h*(2f0 - f0) = h*f0 For f = 5f0: (1/2)m*v2² = h*(5f0 - f0) = 4h*f0 Dividing: v1²/v2² = h*f0 / (4h*f0) = 1/4 v1/v2 = 1/2 Ratio v1: v2 = 1: 2.

Question 48

In an X-ray tube operating at voltage V, the cut-off (minimum) wavelength is lambda0. If the operating voltage is increased slightly by delta_V, which of the following statements is correct?

Accepted Answer

lambda0 - lambda_Kalpha changes by -lambda0 * delta_V / V. lambda0 = hc/(eV), so delta_lambda0 = -(lambda0/V)*delta_V (negative since lambda0 decreases as V increases). lambda_Kalpha is fixed by atomic energy levels. Therefore lambda0 - lambda_Kalpha changes by delta_lambda0 = -lambda0*delta_V/V. The option stating lambda_Kalpha - lambda0 does not change is wrong because lambda0 itself changes.

Question 49

For an electron, if the uncertainty in momentum is twice the uncertainty in position, find the uncertainty in velocity. (Given: h-bar = h/(2*pi))

Accepted Answer

(1/(2m)) * sqrt(h-bar). Let deltaₚ = uncertainty in momentum, deltaₓ = uncertainty in position. Given: deltaₚ = 2*deltaₓ. Heisenberg principle (minimum): deltaₚ * deltaₓ >= h-bar/2. Using equality: deltaₚ * deltaₓ = h-bar/2. Substituting deltaₚ = 2*deltaₓ: 2*(deltaₓ)² = h-bar/2 => deltaₓ = sqrt(h-bar/4) = sqrt(h-bar)/2. deltaₚ = 2*deltaₓ = sqrt(h-bar). delta_v = deltaₚ/m = sqrt(h-bar)/m = (1/m)*sqrt(h-bar). This matches option D. But wait: 'h' in the problem likely means h-bar itself (since the problem says 'Given: h = h/2pi', meaning their symbol h IS h-bar). If their h IS h-bar, then the del

Question 50

A metal surface with work function phi = 2.1 eV is irradiated with photons of energy 4.0 eV. What is the maximum kinetic energy of the photoelectrons emitted?

Accepted Answer

1.9 eV. By Einstein's photoelectric equation, the maximum kinetic energy of an emitted photoelectron equals the incident photon energy minus the work function of the metal. KE_max = E_photon - phi = 4.0 eV - 2.1 eV = 1.9 eV. The work function 2.1 eV is the energy required to release an electron from the surface; the excess 1.9 eV appears as kinetic energy.

JEE Advanced Physics: Dual Nature of Radiation and Matter questions with solutions

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