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JEE Main 20198 AprilMorning Shift

Exam: JEE Main
Year: 2019
Date: 8 April
Shift: Morning Shift
Subject: Physics
Chapter: Atomic Physics

JEE Main 2019 · 8 April · Morning Shift · Physics · Atomic Physics

Radiation coming from transitions n = 2 to n = 1 of hydrogen atoms...

Radiation coming from transitions

n = 2

to

n = 1

of hydrogen atoms fall on

{He}^{+}

ions in

n = 1

and

n = 2

states. The possible transition of helium ions as they absorb energy from the radiation is:

Options

A $n = 2 ⟶ n = 3$
B $n = 2 ⟶ n = 4$
C $n = 2 ⟶ n = 5$
D $n = 1 ⟶ n = 4$

Correct answer

B. $n = 2 ⟶ n = 4$

Step-by-step solution

E = 13.6\left(\dfrac{1}{{1}^{1}} - \dfrac{1}{{2}^{2}}\right)eV . . . ( 1 )

For Helium:

E = 13.6{\left(2\right)}^{2}\left(1{n}_{1}^{2} - 1{n}_{2}^{2}\right)eV . . . ( 2 )

From equation

( 1 )

and

( 2 )

13.6\left(\dfrac{3}{4}\right) = 13.6\left(4\right)\left(1{n}_{1}^{2} - 1{n}_{2}^{2}\right)

1{n}_{1}^{2} - 1{n}_{2}^{2} = \dfrac{3}{16}

From option,

{n}_{1} = 2

and

{n}_{2} = 4

are satisfying the equation.

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