Q: For a proton–electron system, the interaction potential is V = V0 ln(r/r0), where r0 is a constant. If Bohr’s model is assumed to hold, how does the radius rₙ depend on the principal quantum number n?

rₙ is directly proportional to n. With V = V0 ln(r/r0), force F = V0/r. Setting mv^2/r = V0/r gives mv^2 = V0, so v is constant. Bohr quantization mvr = n*hbar then gives r_n ∝ n.

Q: In the hydrogen spectrum, the longest wavelength in the ultraviolet region is 122 nm. The shortest wavelength in the infrared region, rounded to the nearest whole number, is

823 nm. Longest UV is Lyman 2->1: 1/122 = R(3/4) -> R = 1/91.5 nm^-1. Shortest IR is the Paschen series limit (inf->3): lambda = 9/R = 9*91.5 = 823 nm.

Question 1

An electron in one Bohr orbit with principal quantum number n = n1 can produce 3 distinct spectral lines on de-excitation. In another orbit with principal quantum number n = n2, it can produce 6 distinct spectral lines. What is the ratio of the orbital speeds of the electron in these two o

Accepted Answer

4: 3. Number of spectral lines = n(n-1)/2, so 3 lines -> n1 = 3 and 6 lines -> n2 = 4. Orbital speed v is proportional to 1/n, so v1:v2 = (1/3):(1/4) = 4:3.

Question 2

An electron in a hydrogen atom is found in an excited state with energy equal to -3.4 eV. What is its angular momentum?

Accepted Answer

2.10 × 10⁻³⁴ Js. E = -13.6/n^2 = -3.4 eV gives n = 2. Angular momentum L = n*hbar = 2 * 1.055 x 10^-34 = 2.11 x 10^-34 Js.

Question 3

A beam of electrons having energy 12.5 eV is directed at hydrogen gas in the gaseous state at room temperature. The emitted spectrum will contain:

Accepted Answer

2 lines in the Lyman series and 1 line in the Balmer series. From n=1, exciting to n=3 needs 12.09 eV (possible) but to n=4 needs 12.75 eV (>12.5 eV, not possible). Allowed transitions are 3->1, 2->1 (Lyman, 2 lines) and 3->2 (Balmer, 1 line): 2 Lyman + 1 Balmer.

Question 4

A hydrogen atom and a Li++ ion are each in their second excited state. If their electronic angular momenta are denoted by L_H and L_Li, and their energies by E_H and E_Li respectively, which statement is correct?

Accepted Answer

L_H = L_Li and |E_H| < |E_Li|. Both are in the second excited state (n=3), so angular momentum L = nh/2pi is the same: L_H = L_Li. Energy |E| = 13.6 Z^2/n^2, and Li++ has Z=3 vs Z=1 for H, so |E_H| < |E_Li|.

Question 5

For a proton–electron system, the interaction potential is V = V0 ln(r/r0), where r0 is a constant. If Bohr’s model is assumed to hold, how does the radius rₙ depend on the principal quantum number n?

Accepted Answer

rₙ is directly proportional to n. With V = V0 ln(r/r0), force F = V0/r. Setting mv^2/r = V0/r gives mv^2 = V0, so v is constant. Bohr quantization mvr = n*hbar then gives r_n ∝ n.

Question 6

In the hydrogen spectrum, the longest wavelength in the ultraviolet region is 122 nm. The shortest wavelength in the infrared region, rounded to the nearest whole number, is

Accepted Answer

823 nm. Longest UV is Lyman 2->1: 1/122 = R(3/4) -> R = 1/91.5 nm^-1. Shortest IR is the Paschen series limit (inf->3): lambda = 9/R = 9*91.5 = 823 nm.

Question 7

The ionization energy of a hydrogen atom is 13.6 eV. If hydrogen atoms initially in the ground state are exposed to monochromatic light of photon energy 12.1 eV, then as per Bohr’s model, how many spectral lines can be produced in the emitted radiation?

Accepted Answer

three. Absorbing 12.1 eV from the ground state lifts the atom to E = -13.6 + 12.1 = -1.5 eV, i.e. n = 3. De-excitation from n = 3 gives 3(3-1)/2 = 3 spectral lines.

Question 8

An electron in a hydrogen atom makes a transition from the third excited level to the second excited level, and then from the second excited level to the first excited level. What is the ratio of the emitted wavelengths λ₁: λ₂ in these two transitions?

Accepted Answer

20/7. The ratio of emitted wavelengths in transitions between energy levels in a hydrogen atom is inversely proportional to the energy differences between those levels. The energy difference for each transition can be calculated using the Rydberg formula, leading to the ratio of wavelengths being 20/7 for the specified transitions.

Question 9

For a helium ion (He+) in its third Bohr orbit, using the non-relativistic model, what is the electron’s speed? Take K = 9 × 10⁹ kg m²/C², Z = 2, and Planck’s constant h = 6.6 × 10⁻³⁴ J s.

Accepted Answer

1.46 × 10⁶ m/s. Orbital speed v_n = (Z/n)*2.18x10^6 m/s. For He+ with Z = 2, n = 3: v = (2/3)*2.18x10^6 ~ 1.46x10^6 m/s.

Question 10

In the hydrogen spectrum, the Hα emission line has a wavelength of 656 nm. In the spectrum of a faraway galaxy, the same Hα line is observed at 706 nm. The approximate speed of the galaxy relative to the Earth is

Accepted Answer

2 × 10⁷ m/s. Redshift delta lambda = 706 - 656 = 50 nm. v = c (delta lambda / lambda) = 3 x 10^8 x (50/656) = 2.3 x 10^7 m/s ~ 2 x 10^7 m/s.

Question 11

Let ν1 denote the frequency at the Lyman series limit, ν2 the frequency of the first Lyman line, and ν3 the frequency at the Balmer series limit. Which relation is correct?

Accepted Answer

ν1 − ν2 = ν3. nu1 = Rc (Lyman limit), nu2 = Rc(1 - 1/4) = (3/4)Rc (first Lyman line), nu3 = Rc/4 (Balmer limit). Then nu1 - nu2 = Rc/4 = nu3.

Question 12

If the ground-state energy of He+ is -54.4 eV, what is the energy of Li2+ in its first excited state?

Accepted Answer

-30.6 eV. E = -13.6*Z^2/n^2 eV. For Li2+ (Z = 3) in the first excited state (n = 2): E = -13.6*9/4 = -30.6 eV.

Question 13

A spectral line in the emission spectrum of Li2+ has the same wavelength as the second line of the Balmer series of hydrogen. If this Li2+ line arises from the transition n = 12 to n = x, what is the value of x?

Accepted Answer

6. Second Balmer line of H is 4->2: 1/lambda = R(1/4 - 1/16) = 3R/16. For Li2+ (Z=3): 1/lambda = 9R(1/x^2 - 1/12^2). Setting 9(1/x^2 - 1/144) = 3/16 gives 1/x^2 = 1/36, so x = 6.

Question 14

An electron is pulled toward the origin by a central force of magnitude k/r, where k is a constant and r is the electron’s distance from the origin. If the Bohr model is used for this system, the radius of the nth orbit is rₙ and the electron’s kinetic energy is Tₙ. Which statement is corr

Accepted Answer

Tₙ does not depend on n, and rₙ varies as n. For F = k/r, mv^2/r = k/r gives mv^2 = k (constant), so kinetic energy T_n = k/2 is independent of n. Bohr quantization mvr = n*hbar then gives r_n ∝ n.

Question 15

A neutral helium atom needs 24.6 eV to eject one electron. What energy, in eV, is needed to remove both electrons from the neutral helium atom?

Accepted Answer

79.0. To remove both electrons from a neutral helium atom, you need to account for the energy required to remove the first electron (24.6 eV) and the increased energy needed to remove the second electron due to the positive charge of the helium ion formed after the first electron is removed. The total energy required is 24.6 eV for the first electron plus 54.4 eV for the second, resulting in 79.0 eV.

Question 16

When an electron in a hydrogen atom drops from an excited level to the ground level, which statement is correct?

Accepted Answer

Its kinetic energy rises, while its potential energy and total energy both fall.. Dropping to a lower n makes total energy more negative (falls) and KE = -E_total increases (rises), while PE = 2*E_total becomes more negative (falls). So KE rises while PE and total energy both fall.

Question 17

The formation of covalent bonds in compounds demonstrates the:

Accepted Answer

wave character of the electron. Covalent bonds involve the sharing of electron pairs between atoms, which is best described by the wave nature of electrons as it allows for the overlapping of atomic orbitals, leading to stable bond formation.

Question 18

An electron is pulled toward the origin by a central force of magnitude k/r, where k is a constant and r is the electron’s distance from the origin. If the Bohr model is used for this system, the radius of the nth allowed orbit is rₙ and the electron’s kinetic energy is Tₙ. Which relation

Accepted Answer

Tₙ does not depend on n, and rₙ varies as n. In the Bohr model, the kinetic energy of the electron in a hydrogen-like atom is constant for all orbits, meaning it does not depend on the principal quantum number n. However, the radius of the allowed orbits increases linearly with n, leading to the conclusion that Tₙ remains constant while rₙ varies with n.

Question 19

The transition from the state n = 4 to n = 3 in a hydrogen like atom results in ultraviolet radiation. Infrared radiation will be obtained for the transition from:

Accepted Answer

5 → 4. Transition energy scales with (1/nf^2 - 1/ni^2). For 4->3 it is 0.0486 (UV). The only transition with a SMALLER gap (longer wavelength = IR) is 5->4 (1/16 - 1/25 = 0.0225). 3->2, 4->2 and 2->1 all have larger gaps, so they are more energetic, not IR.

Question 20

Energy required for the electron excitation in Li++ from the first to the third Bohr orbit is:

Accepted Answer

108.8 eV. For Li++ (Z=3): E = 13.6*Z^2*(1/1^2 - 1/3^2) = 13.6*9*(8/9) = 108.8 eV.

Question 21

Hydrogen atom is excited from ground state to another state with principal quantum number equal to 4. Then the number of spectral lines in the emission spectra will be:

Accepted Answer

6. From an excited state n = 4 down to ground, the number of distinct spectral lines is n(n-1)/2 = 4*3/2 = 6.

Question 22

In a hydrogen like atom electron make transition from an energy level with quantum number n to another with quantum number (n−1). If n>>1, the frequency of radiation emitted is proportional to:

Accepted Answer

1/n³. E_n = -RhcZ^2/n^2, so for n -> (n-1) the energy gap ~ 2Rhc/n^3 for large n. Hence the emitted frequency is proportional to 1/n^3.

Question 23

Hydrogen (¹H), Deuterium (¹H²), singly ionised Helium (²He⁴⁺), and doubly ionised lithium (³Li⁶⁺) all have one electron around the nucleus. Consider an electron transition from n = 2 to n = 1. If the wavelengths of emitted radiation are λ₁, λ₂, λ₃ and λ₄ respectively then approximately whi

Accepted Answer

λ₁ = λ₂ = 4λ₃ = 9λ₄. For n=2->1, 1/lambda is proportional to Z^2*(3/4), so lambda ~ 1/Z^2. H and D both have Z=1 (same wavelength), He+ has Z=2 (1/4), Li2+ has Z=3 (1/9). Hence lambda1 = lambda2 = 4*lambda3 = 9*lambda4.

Question 24

As an electron makes a transition from an excited state to the ground state of a hydrogen - like atom/ion:

Accepted Answer

its kinetic energy increases but potential energy and total energy decrease. As the electron transitions to a lower energy state, it moves closer to the nucleus, which increases its kinetic energy due to the attractive force. However, the potential energy decreases because it is at a lower energy level, resulting in a decrease in total energy.

Question 25

An electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let λₙ, λ_B be the Broglie wavelength of the electron in the nth state and the wavelength of the emitted photon in the transition from the nth state to the ground state. Let λₙ ∝ n, (A, B a

Accepted Answer

Λₙ ≈ A + B/λₙ². For the n->1 transition, 1/Lam_n = R(1 - 1/n^2), so Lam_n = 1/[R(1-1/n^2)] ~= (1/R)(1 + 1/n^2). Since the de Broglie wavelength lam_n is proportional to n, 1/n^2 is proportional to 1/lam_n^2, giving Lam_n ~= A + B/lam_n^2.

Question 26

If the series limit frequency of the Lyman series is ν_L, then the series limit frequency of the P-fund series is:

Accepted Answer

ν_L/25. The series limit frequency of the P-fund series is derived from the energy levels of the hydrogen atom, specifically the transition from the n=5 level to the ionization limit, which results in a frequency that is 1/25th of the series limit frequency of the Lyman series, which corresponds to transitions from n=2 to the ionization limit.

Question 27

Taking the wavelength of first Balmer line in hydrogen spectrum (n = 3 to n = 2) as 660 nm, the wavelength of the 2nd Balmer line (n = 4 to n = 2) will be:

Accepted Answer

488.9 nm. The wavelength of the second Balmer line can be calculated using the Rydberg formula for hydrogen, which shows that the wavelengths decrease as the initial energy level increases. Since the first Balmer line corresponds to a transition from n=3 to n=2, the second line from n=4 to n=2 will have a shorter wavelength, which is correctly calculated to be 488.9 nm.

Question 28

The time period of revolution of electron in its ground state orbit in a hydrogen atom is 1.6 × 10⁻¹⁶ s. The frequency of revolution of the electron in its first excited state (in s⁻¹) is:

Accepted Answer

7.8 × 10¹⁴. The frequency of revolution is inversely proportional to the time period of revolution. Since the time period for the ground state is given, we can use the relationship between energy levels and their corresponding frequencies to determine that the first excited state has a shorter time period, resulting in a higher frequency of 7.8 × 10¹⁴ s⁻¹.

Question 29

In a hydrogen like atom an electron makes transition from an energy level with quantum number n to another with quantum number (n - 1). If n >> 1, the frequency of radiation emitted is proportional to -

Accepted Answer

1/n³. The frequency of radiation emitted during a transition between energy levels in a hydrogen-like atom is determined by the difference in energy levels, which is inversely proportional to the square of the principal quantum number. For large n, this relationship simplifies to being proportional to 1/n³.

Question 30

Energy of an electron is given by E = -2.178 × 10⁻¹⁸ J (Z²/n²). Wavelength of light required to excite an electron in an hydrogen atom from level n = 1 to n = 2 will be: (h = 6.62 × 10⁻³⁴ Js and c = 3.0 × 10⁸ m s⁻¹)

Accepted Answer

1.214 × 10⁻⁷ m. The energy difference between the two levels (n=1 and n=2) in a hydrogen atom can be calculated using the formula provided, and then the wavelength can be determined using the relationship between energy and wavelength (E = hc/λ). The correct option corresponds to the wavelength that matches the calculated energy difference for the transition.

Question 31

Hydrogen (¹H¹), Deuterium (¹H²), singly ionised Helium (²He⁴⁺) and doubly ionised lithium (³Li⁶⁺) all have one electron around the nucleus. Consider an electron transition from n = 2 to n = 1. If the wavelengths of emitted radiation are λ1, λ2, λ3 and λ4 respectively then approximately whi

Accepted Answer

λ1 = λ2 = 4λ3 = 9λ4. The wavelengths of emitted radiation during electron transitions are inversely proportional to the energy levels of the electrons in the atoms. Since the energy levels depend on the atomic number and the electron configuration, the ratios of the wavelengths can be derived from the differences in these energy levels for each atom. In this case, the correct option reflects the relationship between the wavelengths based on the respective atomic structures and their transitions.

Question 32

As an electron makes a transition from an excited state to the ground state of a hydrogen-like atom from:
(1) its kinetic energy increases but potential energy and total energy decrease
(2) kinetic energy, potential energy and total energy decrease
(3) kinetic energy decreases, potential e

Accepted Answer

its kinetic energy increases but potential energy and total energy decrease. When an electron transitions from an excited state to the ground state, it releases energy, which results in a decrease in total energy. As it moves closer to the nucleus, its potential energy becomes more negative (increases in magnitude), while its kinetic energy increases due to the conservation of energy.

Question 33

The energy required to remove the electron from a singly ionized Helium atom is 2.2 times the energy required to remove an electron from Helium atom. The total energy required to ionize the Helium atom completely is:

Accepted Answer

79 eV. The energy to remove the first electron from a neutral Helium atom is approximately 24.6 eV. Since the energy required to remove the second electron from the singly ionized Helium atom is 2.2 times that, it amounts to about 54.6 eV. Adding these two values gives a total of approximately 79 eV for complete ionization.

Question 34

An electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let λₙ, λ_b be the de Broglie wavelength of the electron in the n^th state and the ground state respectively. Let Λₙ be the wavelength of the emitted photon in the transition from the n^th

Accepted Answer

Λₙ ≈ A + B/λₙ². The correct option indicates that the wavelength of the emitted photon, Λₙ, is inversely related to the square of the de Broglie wavelength of the electron in the n^th state, which aligns with the principles of quantum mechanics where higher energy transitions result in shorter wavelengths of emitted radiation.

Question 35

In a hydrogen like atom, when an electron jumps from the M-shell to the L-shell, the wavelength of emitted radiation is λ. If an electron jumps from N-shell to the L-shell, the wavelength of emitted radiation will be:

Accepted Answer

20/27 λ. For M->L (3->2): 1/lambda = R(1/4 - 1/9) = 5R/36. For N->L (4->2): 1/lambda' = R(1/4 - 1/16) = 3R/16. So lambda'/lambda = (5/36)/(3/16) = 80/108 = 20/27. The emitted wavelength is (20/27) lambda.

Question 36

For emission line of atomic hydrogen from n_i = 8 to n_f = n, the plot of wave number (̅ν) against (1/n²) will be (The Rydberg constant, R_H is wave number unit)
(1) Linear with slope -R_H
(2) Linear with slope R_H
(3) Non linear
(4) Linear with intercept -R_H

Accepted Answer

Linear with slope R_H. Wave number nu-bar = R_H(1/n_f^2 - 1/n_i^2) = R_H(1/n^2) - R_H/64. Plotting nu-bar against (1/n^2) gives a straight line of slope +R_H and intercept -R_H/64, so it is linear with slope R_H.

Question 37

Consider an electron in a hydrogen atom revolving in its second excited state (having radius 4.65 Å). The de-Broglie wavelength of this electron is:

Accepted Answer

9.7 Å. Second excited state means n=3, r=4.65 A. de-Broglie wavelength lambda = 2*pi*r/n = 2*pi*4.65/3 = 9.7 A. So the answer is 9.7 A.

Question 38

The electron in a hydrogen atom first jumps from the third excited state to the second excited state and subsequently to the first excited state. The ratio of the respective wavelengths λ1/λ2 of the photons emitted in this process is:

Accepted Answer

20/7. 3rd excited (n=4) -> 2nd excited (n=3): 1/lambda1 = R(1/9 - 1/16) = 7R/144. 2nd excited (n=3) -> 1st excited (n=2): 1/lambda2 = R(1/4 - 1/9) = 5R/36. lambda1/lambda2 = (5/36)/(7/144) = 20/7.

Question 39

A He⁺ ion is in its first excited state. Its ionization energy is:

Accepted Answer

13.60 eV. The ionization energy of a He⁺ ion in its first excited state corresponds to the energy required to remove an electron from that state, which is calculated using the formula for hydrogen-like atoms. For He⁺, the ionization energy from the ground state is 54.4 eV, and the first excited state corresponds to 13.6 eV, making option B the correct answer.

Question 40

Which one of the following about an electron occupying the 1s orbital in a hydrogen atom is incorrect ? (The Bohr radius is represented by a0).
(1) The magnitude of the potential energy is double that of its kinetic energy on an average.
(2) The probability density of finding the electron

Accepted Answer

The total energy of the electron is maximum when it is at a distance a0 from the nucleus.. For the 1s electron, virial theorem (|PE|=2*KE), max probability density at the nucleus, and finite probability at 2a0 are all correct. The incorrect statement is that total energy is maximum at distance a0 -- the total (orbital) energy is constant and independent of position, so option (3) is the wrong statement.

Question 41

A hydrogen atom, initially in the ground state is excited by absorbing a photon of wavelength 980Å. The radius of the atom in the excited state, in terms of Bohr radius a0 will be: [hc = 12500 eV Å]

Accepted Answer

16a0. Photon energy = 12500/980 = 12.76 eV. From n=1 (-13.6 eV) the atom reaches E = -13.6 + 12.76 = -0.85 eV = -13.6/n^2, giving n^2 = 16, n = 4. Radius = n^2 * a0 = 16 a0.

Question 42

Heat treatment of muscular pain involves radiation of wavelength of about 900 nm. Which spectral line of H-atom is suitable for this purpose? [R_H = 1 × 10⁵ cm⁻¹, h = 6.6 × 10⁻³⁴ J s, c = 3 × 10⁸ m/s]

Accepted Answer

Paschen, ∞ → 3. For Paschen series limit (infinity -> 3): 1/lambda = R_H/9 = (1e5 cm^-1)/9, giving lambda = 9e-5 cm = 900 nm. So the suitable line is Paschen, infinity -> 3. (Balmer infinity -> 2 gives 400 nm.)

Question 43

The shortest wavelength of H atoms in the Lyman series is λ₁. The longest wavelength in the Balmer series of He⁺ is:

Accepted Answer

9λ₁/5. Shortest Lyman wavelength of H: 1/lambda1 = R(1) so lambda1 = 1/R. Longest Balmer line of He+ (Z=2) is 3->2: 1/lambda = R*4*(1/4 - 1/9) = 5R/9, so lambda = 9/(5R) = (9/5)lambda1.

Question 44

The difference between the radii of 3rd and 4th orbits of Li2+ is ΔR1. The difference between the radii of 3rd and 4th orbits of He+ is ΔR2. ΔR1: ΔR2 is:

Accepted Answer

2: 3. Radius r = 0.529 n^2/Z. ΔR(3->4) = (16-9)/Z = 7/Z. For Li2+ (Z=3): ΔR1 = 7/3; for He+ (Z=2): ΔR2 = 7/2. ΔR1:ΔR2 = (7/3):(7/2) = 2:3.

Question 45

The time period of revolution of electron in its ground state orbit in a hydrogen atom is 1.6 × 10⁻¹⁶ s. The frequency of revolution of the electron in its first excited state (in s⁻¹) is -

Accepted Answer

7.8 × 10¹⁴. The frequency of revolution is inversely proportional to the time period of revolution. Since the time period for the ground state is known, we can calculate the time period for the first excited state, which is longer due to the electron being further from the nucleus, resulting in a lower frequency. The correct option reflects this calculation, yielding a frequency of 7.8 × 10¹⁴ s⁻¹.

Question 46

The energy required to ionize a hydrogen like ion in its ground state is 9 Rydbergs. What is the wavelength of the radiation emitted when the electron in this ion jumps from the second excited state to the ground state ?

Accepted Answer

11.4 nm. Ionization energy = 9 Ry gives Z^2 = 9. Energy of n=3 -> n=1 transition = Z^2 * 13.6 * (1 - 1/9) = 9*13.6*(8/9) = 108.8 eV. Wavelength = 1240/108.8 = 11.4 nm.

Question 47

Which level of the single ionized carbon has the same energy as the ground state energy of hydrogen atom ?

Accepted Answer

6. For a one-electron ion E = -13.6 Z^2/n^2 eV. Carbon has Z = 6, so E = -13.6*36/n^2. Setting this equal to the hydrogen ground state (-13.6 eV) gives n^2 = 36, i.e. n = 6.

Question 48

The atomic hydrogen emits a line spectrum consisting of various series. Which series of hydrogen atomic spectra is lying in the visible region?

Accepted Answer

Balmer series. The Balmer series corresponds to electron transitions from higher energy levels to the second energy level, resulting in the emission of visible light, which is why it is the only series among the options that lies in the visible region of the spectrum.

Question 49

Imagine that the electron in a hydrogen atom is replaced by a muon (μ). The mass of muon particle is 207 times that of an electron. The ionization potential of this hydrogen atom will be:-

Accepted Answer

2815.2 eV. For a muonic hydrogen atom the ionization energy = 13.6 eV times the mass ratio = 13.6 x 207 = 2815.2 eV (the muon is ~207 times heavier, and energy is proportional to mass).

Question 50

A certain orbital has no angular nodes and two radial nodes. The orbital is:

Accepted Answer

3s. No angular nodes means l = 0 (an s orbital). Radial nodes = n - l - 1 = 2 gives n = 3. Hence the orbital is 3s.

JEE Main Physics: Atoms questions with solutions

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