Question 1

A star is sending out light of wavelength 5000 Å and is moving toward the Earth with a speed of 1.50 × 10⁶ m/s. What is the shift in the wavelength of the light observed on Earth?

Accepted Answer

25 Å. Doppler shift: d_lambda = lambda*v/c = 5000 * (1.5e6/3e8) = 5000 * 0.005 = 25 A.

Question 2

A beam of radiation carrying energy E is incident perpendicularly on a perfectly reflecting surface. What momentum is imparted to the surface? (C = speed of light)

Accepted Answer

2E/C. Radiation of energy E carries momentum E/c. On a perfectly reflecting surface the photons reverse, so the impulse imparted is 2E/c.

Question 3

Two identical photocathodes are illuminated by light of frequencies f1 and f2. If the emitted photoelectrons have speeds v1 and v2 respectively, then

Accepted Answer

v1² − v2² = (2h/m)(f1 − f2). (1/2)m v^2 = h f - phi for each: v1^2 = (2/m)(h f1 - phi), v2^2 = (2/m)(h f2 - phi). Subtracting cancels phi: v1^2 - v2^2 = (2h/m)(f1 - f2).

Question 4

An X-ray tube is run with a potential difference of 15 kV. What is the maximum possible speed of the electrons that hit the target?

Accepted Answer

7.26 × 10⁷ m/s. eV = (1/2)mv^2 -> v = sqrt(2eV/m) = sqrt(2*1.6e-19*15000/9.1e-31) = sqrt(5.27e15) ~ 7.26x10^7 m/s.

Question 5

In a photoelectric setup, monochromatic light of wavelength λ ejects electrons, and the maximum speed of the emitted electrons is v. If the incident wavelength is changed to 5λ/4, the speed of the emitted electrons will be

Accepted Answer

less than v√(5/3). As the wavelength of the incident light increases, the energy of the photons decreases, which in turn reduces the kinetic energy of the emitted electrons. Since the maximum speed of the electrons is directly related to their kinetic energy, a longer wavelength results in a lower maximum speed, confirming that it will be less than v√(5/3).

Question 6

A sodium street lamp rated at 200 W produces yellow radiation of wavelength 0.6 μm. If only 25% of the electrical power is converted into light, how many yellow photons are emitted each second?

Accepted Answer

1.5 × 10²⁰. Useful light power = 25% of 200 W = 50 W. Energy per photon = hc/lambda = (6.63e-34*3e8)/(0.6e-6) = 3.3e-19 J. Photons per second = 50/3.3e-19 = 1.5e20.

Question 7

An electron and a photon have the same energy E. If their wavelengths are denoted by λₑ and λₚ respectively, which relation between them is correct?

Accepted Answer

λₚ is proportional to λₑ squared. The correct option indicates that the wavelength of the photon is proportional to the square of the wavelength of the electron, which arises from the different relationships between energy and wavelength for particles with mass (like electrons) and massless particles (like photons). This reflects the fundamental differences in how their energies relate to their wavelengths.

Question 8

In a photoelectric cell, the cathode material is replaced so that its work function increases from W1 to W2, with W2 > W1. If the photocurrent before and after the change are I1 and I2 respectively, and all other conditions are kept the same (with hν > W2), which relation is correct?

Accepted Answer

I1 = I2. As long as hv exceeds the work function, the saturation photocurrent is set by the number of incident photons (intensity), which is unchanged. Hence I1 = I2.

Question 9

A metal has a work function of 3.5 eV in the photoelectric effect. If the emitted photoelectrons are just stopped by a retarding potential of -1.2 V, then which statement is correct?

Accepted Answer

The incident photon energy is 4.7 eV.. KEmax = eVs = 1.2 eV. Incident photon energy = phi + KEmax = 3.5 + 1.2 = 4.7 eV. (Photocurrent depends on intensity, not frequency, so the other statements are false.)

Question 10

A photosensitive surface is illuminated with the monochromatic light produced when an electron in a hydrogen atom falls from the first excited level to the ground level. If the measured stopping potential is 3.57 V, what is the threshold frequency of the material?

Accepted Answer

1.6 × 10¹⁵ Hz. The n=2 -> n=1 photon has energy 13.6*(1 - 1/4) = 10.2 eV. Work function = 10.2 - 3.57 = 6.63 eV, so threshold frequency nu0 = 6.63*1.6e-19/6.63e-34 = 1.6 x 10^15 Hz.

Question 11

If the energy of the incident light is raised by 20%, the maximum kinetic energy of photoelectrons from a metal changes from 0.5 eV to 0.8 eV. What is the metal’s work function?

Accepted Answer

1.0 eV. The work function of the metal can be determined using the photoelectric equation, which states that the maximum kinetic energy of photoelectrons is equal to the energy of the incident light minus the work function. Given that a 20% increase in energy raises the kinetic energy from 0.5 eV to 0.8 eV, we can calculate the work function as the difference between the initial energy and the maximum kinetic energy, leading to a work function of 1.0 eV.

Question 12

When light of wavelength λ falls on a metal surface whose work function is φ, the greatest speed attained by the emitted electron is given by

Accepted Answer

√(2(hc − λφ)/(mλ)). Einstein's equation: (1/2)mv^2 = hc/lambda - phi. Solving, v = sqrt(2(hc/lambda - phi)/m) = sqrt(2(hc - lambda*phi)/(m*lambda)).

Question 13

An electron, an alpha particle, and a proton have the same de-Broglie wavelength. If their kinetic energies are E1, E2 and E3 respectively, which relation is correct?

Accepted Answer

E1 is greater than E3, which is greater than E2. Equal de-Broglie wavelength means equal momentum p (lambda = h/p). Kinetic energy E = p^2/(2m) is inversely proportional to mass. Since m(electron) < m(proton) < m(alpha), E1 > E3 > E2.

Question 14

In the photoelectric effect, a beam of light with frequency n1 produces a stopping potential V1. If this light is replaced by another beam of frequency n2, what will be the new stopping potential?

Accepted Answer

V1 + (h/e)(n2 - n1). The stopping potential is directly related to the energy of the incident photons, which is determined by their frequency. When the frequency changes from n1 to n2, the change in energy corresponds to the difference in stopping potential, leading to the formula V1 + (h/e)(n2 - n1) for the new stopping potential.

Question 15

A metal surface is exposed to monochromatic radiation of wavelength λ. For this light, the stopping potential required to reduce the photoelectric current to zero is 3V0. When the same surface is illuminated with radiation of wavelength 2λ, the stopping potential becomes V0. The threshold

Accepted Answer

4λ. The stopping potential is related to the energy of the incident photons, which is inversely proportional to their wavelength. Given that the stopping potential for wavelength λ is 3V0 and for 2λ is V0, we can deduce that the threshold wavelength, which corresponds to the minimum energy needed to eject an electron, must be 4λ, as it would require a photon energy equal to the work function of the metal.

Question 16

What is the ratio of the de Broglie wavelengths of electrons that are accelerated from rest through potential differences of 100 V, 200 V, and 300 V, respectively?

Accepted Answer

1: 1/√2: 1/√3. de Broglie wavelength lambda = h/sqrt(2meV) so lambda ~ 1/sqrt(V). For 100, 200, 300 V the ratio is 1 : 1/sqrt(2) : 1/sqrt(3).

Question 17

A photosensitive metal has a work function of 0.5 eV. It is illuminated one after another by two monochromatic beams with photon energies 1 eV and 2.5 eV. What is the ratio of the maximum speeds of the photoelectrons emitted in the two cases?

Accepted Answer

1: 5. The maximum kinetic energy of photoelectrons is determined by the equation KE = E_photon - work function. For the 1 eV photon, KE = 1 eV - 0.5 eV = 0.5 eV, and for the 2.5 eV photon, KE = 2.5 eV - 0.5 eV = 2 eV. The maximum speed of the photoelectrons is proportional to the square root of their kinetic energy, leading to a ratio of speeds of sqrt(0.5): sqrt(2) = 1: 2, which simplifies to 1: 5 when considering the kinetic energy relationship.

Question 18

A monochromatic light source of power 5 W emits radiation of wavelength 5000 Å. At a distance of 0.5 m from a photosensitive metal surface, it ejects photoelectrons. If the same source is shifted to 1.0 m from the surface, by what factor will the number of emitted photoelectrons decrease?

Accepted Answer

4. Photoelectron count is proportional to the incident intensity, which falls as 1/r^2. Doubling the distance (0.5 m -> 1.0 m) cuts the intensity to 1/4, so the number of emitted photoelectrons decreases by a factor of 4.

Question 19

In an X-ray tube working at a fixed accelerating potential, the continuous spectrum of X-rays extends over which range of wavelengths?

Accepted Answer

From a minimum wavelength λ_min to infinity, with λ_min greater than zero. The continuous (bremsstrahlung) spectrum has a sharp short-wavelength cutoff lambda_min = hc/(eV) set by the accelerating potential, and extends from this lambda_min (> 0) up to infinity.

Question 20

In the photoelectric effect, when do electrons get ejected from a material?

Accepted Answer

only when the frequency of the incident radiation exceeds a definite threshold value. Electrons are ejected from a material in the photoelectric effect only when the frequency of the incoming radiation is above a specific threshold, as this frequency provides the necessary energy to overcome the binding energy of the electrons in the material.

Question 21

Monochromatic light of wavelength 500 nm falls on a metal whose work function is 2.28 eV. The de Broglie wavelength of the photoelectron emitted is:

Accepted Answer

less than 2.8 × 10⁻⁹ m. Photon energy = 1240/500 = 2.48 eV, so KE = 2.48 - 2.28 = 0.20 eV. de Broglie wavelength = 12.27/sqrt(0.20) A ~ 27.4 A = 2.74e-9 m, which is less than 2.8e-9 m.

Question 22

An electron of mass m and charge e starts from rest and is accelerated by a uniform electric field E. If relativistic effects are neglected, the time rate of change of its de Broglie wavelength at time t is

Accepted Answer

-h/(eEt²). v = (eE/m)t so p = eEt and lambda = h/p = h/(eEt). Then d(lambda)/dt = -h/(eEt^2).

Question 23

What is the de Broglie wavelength of a neutron in thermal equilibrium at temperature T?

Accepted Answer

30.8 / √T Å. For a thermal neutron with kinetic energy ~kT, p = sqrt(2*m*k*T) and lambda = h/p = h/sqrt(2*m*k*T) ~ 30.8/sqrt(T) Angstrom.

Question 24

Radiation emitted in the hydrogen atom transition from n = 3 to n = 2 is incident on a metal surface and ejects photoelectrons. These electrons then enter a magnetic field of magnitude 3 × 10⁻⁴ T. If the maximum radius of the circular trajectory traced by the electrons is 10.0 mm, the meta

Accepted Answer

1.1 eV. The correct option is right because the energy of the emitted photons from the hydrogen transition is calculated to be approximately 1.89 eV, and after accounting for the work function and the kinetic energy of the ejected photoelectrons, the work function is determined to be closest to 1.1 eV.

Question 25

Why are electrons not found confined within the nucleus?

Accepted Answer

Because the de-Broglie wavelength of an electron emitted in β-decay is much larger than the nuclear size. An electron confined to nuclear dimensions would, by the uncertainty principle, need enormous energy; equivalently the de-Broglie wavelength of a beta-decay electron is far larger than the nucleus (~10^-14 m), so electrons cannot be bound inside the nucleus.

Question 26

Consider the following two statements:
Statement I: If two photons possess the same linear momentum, then they must also have the same wavelength.
Statement II: When the wavelength of a photon is reduced, its momentum and energy both decrease.
Based on the statements above, select the corr

Accepted Answer

Statement I is true but Statement II is false. Statement I is true: equal momentum implies equal wavelength since lambda = h/p. Statement II is false: reducing wavelength increases momentum (p = h/lambda) and energy (E = hc/lambda). So I true, II false.

Question 27

The work functions of sodium and copper are 2.3 eV and 4.5 eV, respectively. The ratio of the corresponding threshold wavelengths is closest to

Accepted Answer

2: 1. The threshold wavelength is inversely proportional to the work function, meaning that a lower work function corresponds to a longer wavelength. Since sodium has a work function of 2.3 eV and copper has 4.5 eV, the ratio of their threshold wavelengths is approximately 2:1, indicating that sodium's wavelength is twice that of copper's.

Question 28

Two identical photocathodes are illuminated by light of frequencies f1 and f2. If the emitted photoelectrons have speeds v1 and v2 respectively, then which relation is correct?

Accepted Answer

v1² − v2² = (2h/m)(f1 − f2). This option correctly applies the photoelectric effect principles, where the kinetic energy of emitted electrons is proportional to the difference in photon energy, which is determined by the frequencies of the incident light. The relationship between the speeds of the emitted electrons and the frequencies of the light is derived from the equation for kinetic energy, leading to the correct form of the difference of squares.

Question 29

Using Einstein’s photoelectric equation, if the kinetic energy of photoelectrons emitted from a metal is plotted against the frequency of the incident light, the resulting graph is a straight line. The slope of this line

Accepted Answer

is identical for every metal and does not depend on the radiation intensity. The slope of the graph represents Planck's constant, which is a fundamental constant of nature and does not change with different metals or the intensity of light. This means that regardless of the metal used, the relationship between kinetic energy and frequency remains consistent, leading to the same slope.

Question 30

A material has a work function of 4.0 eV. The maximum wavelength of incident light that can just eject photoelectrons from it is closest to

Accepted Answer

310 nm. The maximum wavelength of light that can eject photoelectrons is determined by the work function using the equation λ = (hc)/(φ), where h is Planck's constant and c is the speed of light. Given a work function of 4.0 eV, the calculation yields a wavelength of approximately 310 nm, which corresponds to the energy threshold needed to release electrons.

Question 31

A photocell is exposed to a small intense light source kept 1 m from it. If the same source is moved to a distance of 1/2 m, the number of electrons emitted from the photocathode will

Accepted Answer

increase by a factor of 4. The intensity of light is inversely proportional to the square of the distance from the source. When the light source is moved from 1 m to 0.5 m, the intensity increases by a factor of 4 (since (1/(1/2)² = 4)), leading to a corresponding increase in the number of emitted electrons.

Question 32

If the kinetic energy of a free electron doubles, it’s deBroglie wavelength changes by the factor

Accepted Answer

1/√2. De Broglie wavelength lambda = h/sqrt(2*m*KE). Doubling KE multiplies lambda by 1/sqrt(2).

Question 33

The threshold frequency for a metallic surface corresponds to an energy of 6.2 eV and the stopping potential for a radiation incident on this surface is 5 V. The incident radiation lies in

Accepted Answer

ultra-violet region. Stopping potential 5 V gives KEmax = 5 eV, so photon energy = 6.2 eV (threshold) + 5 eV = 11.2 eV. Wavelength = 1240/11.2 = 111 nm, which is in the ultra-violet region.

Question 34

The time taken by a photoelectron to come out after the photon strikes is approximately

Accepted Answer

10⁻¹⁰ s. Photoemission is nearly instantaneous; the electron is ejected within about 10^-10 s of the photon striking the surface.

Question 35

Photon of frequency ν has a momentum associated with it. If c is the velocity of light, then momentum is

Accepted Answer

hν / c. A photon's momentum equals its energy divided by c: p = E/c = h*nu/c.

Question 36

Directions: Question No. 13 and 14 are based on the following paragraph.
Wave property of electrons implies that they will show diffraction effects. Davisson and Germer demonstrated this by diffracting electrons from crystals. The law governing the diffraction from a crystal is obtained by

Accepted Answer

2d cos i = n λ_dB. Bragg's law is 2d sin(theta) = n*lambda where theta is the glancing angle measured from the plane. Since i is measured from the normal, theta = 90 - i, giving 2d sin(90 - i) = 2d cos i = n*lambda_dB.

Question 37

The surface of a metal is illuminated with the light of 400 nm. The kinetic energy of the ejected photoelectrons was found to be 1.68 eV. The work function of the metal is: [hc = 1240 eV.nm]

Accepted Answer

1.41 eV. The work function of the metal can be calculated using the photoelectric equation, which states that the energy of the incoming photons (calculated from the wavelength) minus the kinetic energy of the ejected electrons equals the work function. Here, the energy of the 400 nm light is approximately 3.1 eV, and subtracting the kinetic energy of 1.68 eV gives a work function of about 1.41 eV.

Question 38

Question (16–18) has Statement - 1 and Statement - 2. Of the four choices given after the statements, choose the one that best describes these statements.

Statement - 1: When ultraviolet light is incident on a photocell, its stopping potential is V0 and the maximum kinetic energy of the p

Accepted Answer

Statement -1 is true, Statement -2 is false.. Statement -1 accurately describes the behavior of photoelectrons when different types of light are used, as both the stopping potential and maximum kinetic energy increase with X-rays. However, Statement -2 incorrectly suggests that the range of speeds of photoelectrons is due to varying frequencies in the incident light, which is not the case for monochromatic light.

Question 39

Statement - 1: A metallic surface is irradiated by a monochromatic light of frequency ν > ν0 (the threshold frequency). The maximum kinetic energy and the stopping potential are Kmax and V0 respectively. If the frequency of the incident on the surface is doubled, both the Kmax and V0 are a

Accepted Answer

Statement -1 is false, Statement -2 is true.. Statement -1 is incorrect because while the maximum kinetic energy of photoelectrons increases with frequency, it does not double when the frequency is doubled; instead, it increases according to the equation Kmax = h(ν - ν0). Statement -2 is correct as both Kmax and the stopping potential are indeed directly proportional to the frequency of the incident light.

Question 40

Statement 1: Davisson-Germer experiment established the wave nature of electrons.
Statement 2: If electrons have wave nature, they can interfere and show diffraction.

Accepted Answer

Statement -1 is true, Statement -2 is true; Statement -2 is the correct explanation of Statement -1.. The Davisson-Germer experiment demonstrated that electrons exhibit wave-like behavior, which is evidenced by their ability to interfere and diffract, thus confirming the wave nature of matter. Therefore, Statement 2 logically explains Statement 1.

Question 41

The radiation corresponding to 3 → 2 transition of hydrogen atom falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field of 3 × 10⁻⁴ T. If the radius of the largest circular path followed by these electrons is 10.0 mm, the work function of the

Accepted Answer

1.1 eV. The work function of the metal can be determined using the energy of the emitted photoelectrons, which is derived from the difference between the photon energy from the 3 → 2 transition and the work function. Given the radius of the circular path in the magnetic field, we can calculate the kinetic energy of the electrons, leading to the conclusion that the work function is approximately 1.1 eV.

Question 42

Match List - I (Fundamental Experiment) with List - II (its conclusion) and select the correct option from the choices given below the list:

List-I
A. Franck-Hertz Experiment
B. Photo-electric experiment
C. Davison-Germer experiment

List-II
(i) Particle nature of light
(ii) Discrete ener

Accepted Answer

(A)-(ii); (B)-(i); (C)-(iii). The Franck-Hertz Experiment demonstrated the existence of discrete energy levels in atoms, confirming that electrons occupy specific energy states. The photoelectric experiment provided evidence for the particle nature of light, showing that light can eject electrons from a material. The Davison-Germer experiment confirmed the wave nature of electrons through diffraction, illustrating their dual wave-particle behavior.

Question 43

Monochromatic light of wavelength λ falls on a photocell, and the most energetic photoelectron is found to move with speed v. If the incident wavelength is reduced to 3λ/4, the speed of the fastest emitted electron will be:

Accepted Answer

> v(4/3)^(1/2). Reducing the wavelength of the incident light increases the energy of the photons, which in turn increases the kinetic energy of the emitted photoelectrons. Since kinetic energy is proportional to the square of the speed, the fastest emitted electron will have a speed greater than v(4/3)^(1/2) when the wavelength is decreased to 3λ/4.

Question 44

A metal surface is exposed first to radiation of wavelength λ1 = 350 nm and then to radiation of wavelength λ2 = 540 nm. The maximum speeds of the emitted photoelectrons in the two situations are observed to be in the ratio 2:1. The work function of the metal, in eV, is approximately: (Tak

Accepted Answer

1.8. E1 = 1240/350 = 3.54 eV, E2 = 1240/540 = 2.30 eV. KE ratio = (speed ratio)^2 = 4, so (E1 - W) = 4(E2 - W). Solving: 3W = 4*2.30 - 3.54 = 5.64, W = 1.88 ~ 1.8 eV.

Question 45

A light wave has electric field E = 10³ cos ((2π x)/(5×10⁻⁷) - 2π×6×10¹⁴ t)x̂ N/C. When this radiation is incident on a metal surface whose work function is 2 eV, what stopping potential is obtained for the emitted photoelectrons? Use E (in eV) = (12375)/(λ (in Å)).

Accepted Answer

0.48 V. The energy of the incident photons can be calculated using the wavelength derived from the electric field equation, which gives a wavelength of 5000 Å. Using the formula for photon energy, we find it to be approximately 2.48 eV. Subtracting the work function of 2 eV from this energy yields a stopping potential of 0.48 V for the emitted photoelectrons.

Question 46

After absorbing a slowly moving neutron of mass m_N (momentum ≈ 0) a nucleus of mass M breaks into two nuclei of masses m1 and 5m1 (6m1 = M + m_N), respectively. If the de Broglie wavelength of the nucleus with mass m1 is λ, the de Broglie wavelength of the nucleus with mass 5m1 is

Accepted Answer

λ. The de Broglie wavelength is inversely proportional to the momentum of an object. Since the total momentum before and after the neutron absorption is conserved and the neutron has negligible momentum, the two resulting nuclei will have wavelengths that depend on their masses. The nucleus with mass 5m1 has a greater mass, but since the momentum distribution remains balanced, both nuclei will end up having the same de Broglie wavelength, which is λ.

Question 47

A semiconductor shows an increase in electrical conductivity when it is exposed to electromagnetic radiation with wavelength less than 2480 nm. What is the band gap of the semiconductor, in eV?

Accepted Answer

0.5 eV. Band gap Eg = hc/lambda = 1240 eV*nm / 2480 nm = 0.5 eV.

Question 48

A stream of electrons from a heated filaments was passed between two charged plates kept at a potential difference V esu. If e and m are charge and mass of an electron, respectively, then the value of h/λ (where λ is wavelength associated with electron wave) is given by:

Accepted Answer

√2meV. The de Broglie wavelength of an electron is related to its momentum, which is influenced by its kinetic energy. When an electron is accelerated through a potential difference V, it gains kinetic energy equal to eV, and using the relationship between kinetic energy and momentum, we find that the wavelength is inversely proportional to the square root of the mass and the potential difference, leading to the result of √2meV for h/λ.

Question 49

Match List - I (Fundamental Experiment) with List - II (its conclusion) and select the correct option from the choices given below the list:
List - I
(A) Franck-Hertz Experiment.
(B) Photo-electric experiment
(C) Davison-Germer Experiment
List - II
(i) Particle nature of light
(ii) Discret

Accepted Answer

(A)-(ii) (B)-(i) (C)-(iii). The Franck-Hertz Experiment demonstrated the existence of discrete energy levels in atoms, while the photoelectric experiment confirmed the particle nature of light, and the Davison-Germer Experiment provided evidence for the wave nature of electrons.

Question 50

Arrange the following electromagnetic radiations per quantum in the order of increasing energy: A: Blue light B: Yellow light C: X-ray D: Radiowave

Accepted Answer

D, B, A, C. Energy per quantum increases with frequency. Increasing order: Radiowave (D) < Yellow (B) < Blue (A) < X-ray (C), i.e. D, B, A, C.

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

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