Hydrogen (_1H^1), Deuterium… - JEE Main 2014 Physics PYQ

JEE Main 2014 — Physics Modern Physics

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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 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 ?

Official previous-year question

Held on 6 Apr 2014 · Verified 6 Jul 2026.

Options

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

Did you get this right?

Solution

For radiation, we know that

$\frac{1}{\lambda }=R{Z}^{2}(\frac{1}{{n}_{1}^{2}}-\frac{1}{{n}_{2}^{2}})$

$\therefore \lambda \propto \frac{ 1 }{ {\text{Z}}^{2} }$

${\lambda }_{1} \propto \frac{ 1 }{ 1 }$

${\lambda }_{2} \propto \frac{ 1 }{ 1 }$

${\lambda }_{3} \propto \frac{ 1 }{ 4 }$

${\lambda }_{4} \propto \frac{ 1 }{ \text{9} }$

$\therefore {\lambda }_{1} = {\lambda }_{2} = 4 {\lambda }_{3} = { 9 \lambda }_{4}$

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Options

$4 {\lambda }_{1} = 2 {\lambda }_{2} = 2 {\lambda }_{3} = {\lambda }_{4}$

${\lambda }_{1} = 2 {\lambda }_{2} = 2 {\lambda }_{3} = {\lambda }_{4}$

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

${\lambda }_{1} = { 2 \lambda }_{2} = 3 {\lambda }_{3} = { 4 \lambda }_{4}$

Solution

For radiation, we know that

$\frac{1}{\lambda }=R{Z}^{2}(\frac{1}{{n}_{1}^{2}}-\frac{1}{{n}_{2}^{2}})$

$\therefore \lambda \propto \frac{ 1 }{ {\text{Z}}^{2} }$

${\lambda }_{1} \propto \frac{ 1 }{ 1 }$

${\lambda }_{2} \propto \frac{ 1 }{ 1 }$

${\lambda }_{3} \propto \frac{ 1 }{ 4 }$

${\lambda }_{4} \propto \frac{ 1 }{ \text{9} }$

$\therefore {\lambda }_{1} = {\lambda }_{2} = 4 {\lambda }_{3} = { 9 \lambda }_{4}$

Options

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Solution

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Options

Did you get this right?

Solution

Your note

JEE Main Modern Physics in other years

More JEE Main Modern Physics questions

Other Physics topics for JEE Main

Mechanics

Electromagnetism

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Modern Physics