JEE Challenger
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Bohr Model Dependencies: Radius and Angular Momentum Matching

Comprehension Passage

Answer Q.15 and Q.16 by appropriately matching the lists based on the information given in the paragraph.

Consider the Bohr's model of a one-electron atom where the electron moves around the nucleus. In the following, List-I contains some quantities for the nthn^{\text{th}} orbit of the atom and List-II contains options showing how they depend on nn.

List-IList-II
(I) Radius of the nthn^{\text{th}} orbit(P) ∝n−2\propto n^{-2}
(II) Angular momentum of the electron in the nthn^{\text{th}} orbit(Q) ∝n−1\propto n^{-1}
(III) Kinetic energy of the electron in the nthn^{\text{th}} orbit(R) ∝n0\propto n^0
(IV) Potential energy of the electron in the nthn^{\text{th}} orbit(S) ∝n1\propto n^1
(T) ∝n2\propto n^2
(U) ∝n1/2\propto n^{1/2}

Which of the following options has the correct combination considering List-I and List-II?

Options

A

(I), (P)

B

(I), (T)

Correct
C

(II), (Q)

D

(II), (R)

Step-by-Step Solution

According to Bohr's model for a one-electron hydrogen-like atom with nuclear charge ZZ:

  1. Radius of the nthn^{\text{th}} orbit (rnr_n):
    Centripetal force is provided by the electrostatic attraction:
    mvn2rn=Ze24πϵ0rn2\frac{m v_n^2}{r_n} = \frac{Z e^2}{4\pi\epsilon_0 r_n^2}
    From Bohr's quantization condition for angular momentum:
    mvnrn=nh2π  ⟹  vn=nh2πmrnm v_n r_n = \frac{n h}{2\pi} \implies v_n = \frac{n h}{2\pi m r_n}
    Substituting vnv_n:
    rn=ϵ0h2πmZe2n2  ⟹  rn∝n2r_n = \frac{\epsilon_0 h^2}{\pi m Z e^2} n^2 \implies r_n \propto n^2
    Thus, (I) corresponds to (T).

  2. Angular momentum of the electron in the nthn^{\text{th}} orbit (LnL_n):
    Ln=mvnrn=nh2π  ⟹  Ln∝n1L_n = m v_n r_n = \frac{n h}{2\pi} \implies L_n \propto n^1
    Thus, (II) corresponds to (S).

Checking the given options for Question 15:

  • (A) (I), (P) →\rightarrow Incorrect (rn∝n2r_n \propto n^2, not n−2n^{-2})
  • (B) (I), (T) →\rightarrow Correct (rn∝n2r_n \propto n^2)
  • (C) (II), (Q) →\rightarrow Incorrect (Ln∝n1L_n \propto n^1, not n−1n^{-1})
  • (D) (II), (R) →\rightarrow Incorrect (Ln∝n1L_n \propto n^1, not n0n^0)

Therefore, the correct option is B.