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Atoms · Question 1

JEE Advanced 2026 Paper 1Multiple CorrectMedium📈 Topic in 33% papers+4−1
1

Consider a hydrogen atom with vk, rk, and Kk denoting the velocity, orbital radius and kinetic energy of the electron in the kth orbit, respectively. The electron undergoes a transition from the nth orbit, emitting radiation corresponding to the Lyman series. Considering h to be the Planck's constant and 0 the permittivity of the free space, the correct statement(s) is/are:

  1. Magnitude of change in kinetic energy of electron can be expressed as h4 | nvnrn - v1r1 |.
  2. Magnitude of change in de Broglie wavelength of the electron can be expressed as e24 0 | 1Kn - 1K1 |.
  3. Frequency of the radiation emitted can be expressed as e28 0 h ( 1r1 - 1rn ).
  4. Magnitude of change in total energy of the electron can be expressed as h2 | v1r1 - nvnrn |.

Answer

(1) Magnitude of change in kinetic energy of electron can be expressed as h4 | nvnrn - v1r1 |., (3) Frequency of the radiation emitted can be expressed as e28 0 h ( 1r1 - 1rn ).

Solution

Lyman series: transition n 1. Bohr: mvk rk = kh2 , Kk = e28 0 rk, Ek = -Kk.

(A) Kk = 12mvk2 = 12 kh2 vk rkvk2 = h4 kvkrk, so | K| = h4 | nvnrn - v1r1 | — correct.

(B) k = 2 rkk with rk = e28 0 Kk gives k = e24 0 kKk; | | = e24 0 | 1nKn - 1K1 | — the factor n is missing, incorrect.

(C) h = En - E1 = e28 0 ( 1r1 - 1rn ) — correct.

(D) | E| = | K| = h4 | |, not h2 | | — incorrect.

Answer: (A), (C)

JEE Advanced 2026 Paper 1

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Consider an electron in the n=3 orbit of a hydrogen-like atom with atomic number Z. At absolute temperature T, a neutron having thermal energy kBT has the same de Broglie wavelength as that of this electron. If this temperature is given by T= Z2h2 2a02mNkB, (where h is the Planck's constant, kB is the Boltzmann constant, mN is the mass of the neutron and a0 is the first Bohr radius of hydrogen atom) then the value of is

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A hydrogen atom, initially at rest in its ground state, absorbs a photon of frequency 1 and ejects the electron with a kinetic energy of 10 eV. The electron then combines with a positron at rest to form a positronium atom in its ground state and simultaneously emits a photon of frequency 2. The center of mass of the resulting positronium atom moves with a kinetic energy of 5 eV. It is given that positron has the same mass as that of electron and the positronium atom can be considered as a Bohr atom, in which the electron and the positron orbit around their center of mass. Considering no other energy loss during the whole process, the difference between the two photon energies (in eV) is

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A particle of mass m is moving in a circular orbit under the influence of the central force F(r)=-kr, corresponding to the potential energy V(r)=kr2/2, where k is a positive force constant and r is the radial distance from the origin. According to the Bohr's quantization rule, the angular momentum of the particle is given by L=n , where =h/(2 ), h is the Planck's constant, and n a positive integer. If v and E are the speed and total energy of the particle, respectively, then which of the following expression(s) is(are) correct?

  1. r2=n 1mk
  2. v2=n km3
  3. Lmr2= km
  4. E= n 2 km
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A metal target with atomic number Z = 46 is bombarded with a high energy electron beam. The emission of X-rays from the target is analyzed. The ratio r of the wavelengths of the K -line and the cut-off is found to be r = 2. If the same electron beam bombards another metal target with Z = 41, the value of r will be

  1. 2.53
  2. 1.27
  3. 2.24
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