JEE Advanced 2026 · Paper 1 · Physics Q5

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PhysicsMultiple correct+4 / -1Question 21 of 48

Consider a hydrogen atom with $v_k,r_k$ , and $K_k$ denoting the velocity, orbital radius and kinetic energy of the electron in the $k^{\text{th}}$ orbit, respectively. The electron undergoes a transition from the $n^{\text{th}}$ orbit, emitting radiation corresponding to the Lyman series. Considering $h$ to be the Planck's constant and $\epsilon_0$ the permittivity of the free space, the correct statement(s) is/are:

Magnitude of change in kinetic energy of electron can be expressed as

\dfrac h{4\pi}\left|\dfrac{nv_n}{r_n}-\dfrac{v_1}{r_1}\right|

Magnitude of change in de Broglie wavelength of the electron can be expressed as

\dfrac{e^2}{4\epsilon_0}\left|\dfrac1{K_n}-\dfrac1{K_1}\right|

Frequency of the radiation emitted can be expressed as

\dfrac{e^2}{8\pi\epsilon_0h}\left(\dfrac1{r_1}-\dfrac1{r_n}\right)

Magnitude of change in total energy of the electron can be expressed as

\dfrac h{2\pi}\left|\dfrac{v_1}{r_1}-\dfrac{nv_n}{r_n}\right|

Answer and worked solution

Official answer: A, C

Bohr relations. $mv_kr_k=\dfrac{kh}{2\pi}$ , so $mv_k=\dfrac{kh}{2\pi r_k}$ and
$K_k=\tfrac12(mv_k)v_k=\frac{kh}{4\pi}\frac{v_k}{r_k}.$
Also $K_k=\dfrac{e^2}{8\pi\epsilon_0r_k}$ and $E_k=-K_k$ .

(A) $|K_n-K_1|=\dfrac h{4\pi}\left|\dfrac{nv_n}{r_n}-\dfrac{v_1}{r_1}\right|$ . True.

(D) $|\Delta E|=|\Delta K|$ , which has $\frac h{4\pi}$ , not $\frac h{2\pi}$ . False.

(C) $h\nu=E_n-E_1=\dfrac{e^2}{8\pi\epsilon_0}\left(\dfrac1{r_1}-\dfrac1{r_n}\right)$ . True.

(B) $\lambda_k=\dfrac{2\pi r_k}{k}$ and $r_k=\dfrac{e^2}{8\pi\epsilon_0K_k}$ , so $\lambda_k=\dfrac{e^2}{4\epsilon_0\,kK_k}$ . Then $\Delta\lambda=\dfrac{e^2}{4\epsilon_0}\left|\dfrac1{nK_n}-\dfrac1{K_1}\right|$ ; the stated form lacks the factor $\frac1n$ . False.

Topics:Bohr model and hydrogen spectrum de Broglie waves

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JEE Advanced 2026 · Paper 1, Physics question 5