Start free

Chemistry · Structure of Atom

JEE Advanced 2019 — Paper 2 — Question 36

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 nth \mathrm{n}^{\text {th }} orbit of the atom and List-II contains options showing how they depend on nn.

LIST-ILIST-II
(I) Radius of the nth \mathrm{n}^{\text {th }} orbit(P) ∝n−2 \propto n^{-2}
(II) Angular momentum of the electron in the nth \mathrm{n}^{\text {th }} orbit(Q) ∝n−1 \propto n^{-1}
(III) Kinetic energy of the electron in the nth n^{\text {th }} orbit(R) ∝n0 \propto n^{0}
(IV) Potential energy of the electron in the nth n^{\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?

  1. Option A:

    (III), (P)

    Correct
  2. Option B:

    (IV), (U)

  3. Option C:

    (III), (S)

  4. Option D:

    (IV), (Q)

Answer: A

Step-by-step solution

K.E=13.6Z2n2 eV/atom\text{K.E} = \frac{13.6 Z^2}{n^2} \text{ eV/atom} PE=−2×13.6Z2n2 eV/atom\text{PE} = \frac{-2 \times 13.6 Z^2}{n^2} \text{ eV/atom} Radius=0.529n2ZA˚\text{Radius} = 0.529 \frac{n^2}{Z} Å Angular momentum of electron (mvr)=nh2π\text{Angular momentum of electron (mvr)} = \frac{nh}{2\pi}

Answer key and solution verified before publishing.

Practise Structure of Atom

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free
Exam
JEE Advanced 2019
Paper
Paper 2
Subject
Chemistry
Chapter
Structure of Atom
Topic
Bohr's Model of Atom

More Structure of Atom questions from this paper

Consider the Bohr's model of a one-electron atom where the electron… | JEE Advanced 2019 PYQ with Solution · DhiX AI