Hydrogen ( 1 H 1 ), Deuterium ( 1 H 2 ), singly ionised Helium ( 2 He 4 ) + and doubly ionised lithium ( 3 Li 6 ) ++ all have one electron around the nucleus. Consider an electron transition from n = 2 to n = 1 . If the

Correct answer: C. λ _ 1 = λ _ 2 = 4 λ _ 3 = 9 λ _ 4 For radiation, we know that (1)/(λ) = R Z ^ 2 (1 n _ 1 ^ 2 - 1 n _ 2 ^ 2 ) ∴ λ ∝ 1 Z ^ 2 λ _ 1 ∝ (1)/(1) λ _ 2 ∝ (1)/(1) λ _ 3 ∝ (1)/(4) λ _ 4 ∝ 1 9 ∴ λ _ 1 = λ _ 2 = 4 λ _ 3 = 9 λ _ 4

JEE Main 20146 April

Exam: JEE Main
Year: 2014
Date: 6 April
Subject: Physics
Chapter: Atomic Physics

JEE Main 2014 · 6 April · Physics · Atomic Physics

Hydrogen ( 1 H 1 ), Deuterium ( 1 H 2 ), singly ionised Helium ( 2...

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

n = 1

. If the wave lengths of emitted radiation are

{\lambda}_{1} \text{,} {\lambda}_{2} \text{,} {\lambda}_{3}

and

{\lambda}_{4}

respectively then approximately which one of the following is correct ?

Options

A $4 {λ}_{1} = 2 {λ}_{2} = 2 {λ}_{3} = {λ}_{4}$
B ${λ}_{1} = 2 {λ}_{2} = 2 {λ}_{3} = {λ}_{4}$
C ${λ}_{1} = {λ}_{2} = 4 {λ}_{3} = {9 λ}_{4}$
D ${λ}_{1} = {2 λ}_{2} = 3 {λ}_{3} = {4 λ}_{4}$

Correct answer

C. ${λ}_{1} = {λ}_{2} = 4 {λ}_{3} = {9 λ}_{4}$

Step-by-step solution

For radiation, we know that

\dfrac{1}{\lambda} = R{Z}^{2}\left(1{n}_{1}^{2} - 1{n}_{2}^{2}\right)

∴ \lambda ∝ \dfrac{1}{{\text{Z}}^{2}}

{\lambda}_{1} ∝ \dfrac{1}{1}

{\lambda}_{2} ∝ \dfrac{1}{1}

{\lambda}_{3} ∝ \dfrac{1}{4}

{\lambda}_{4} ∝ \dfrac{1}{\text{9}}

∴ {\lambda}_{1} = {\lambda}_{2} = 4 {\lambda}_{3} = {9 \lambda}_{4}

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