15 eV is given to e^ - in 4^ th orbit, then find its final energy when it comes out of H -atom.

Correct answer: A. 14.15 eV Energy of 4^ th orbit of H -atom =-13.6 × (1)/(16)=-0.85 eV So energy released = total energy - ionization energy of 4^ th orbit =15-0.85=14.15 eV

AIIMS 201926 May

Exam: AIIMS
Year: 2019
Date: 26 May
Subject: Physics
Chapter: Atomic Physics

AIIMS 2019 · 26 May · Physics · Atomic Physics

15 eV is given to e^ - in 4^ th orbit, then find its final energy...

15 \mathrm{eV} is given to

e^{-}

in

4^{

\text{th }

}

orbit, then find its final energy when it comes out of

\mathrm{H}

-atom.

Options

A $14.15 \mathrm{eV}$
B $13.6 \mathrm{eV}$
C $12.08 \mathrm{eV}$
D $15.85 \mathrm{eV}$

Correct answer

A. $14.15 \mathrm{eV}$

Step-by-step solution

Energy of

4^{\text{th }}

orbit of

\mathrm{H}

-atom

=-13.6 \times \frac{1}{16}=-0.85 \mathrm{eV}

So energy released = total energy - ionization energy of

4^{\text{th }}

orbit

=15-0.85=14.15 \mathrm{eV}

Practice Atomic Physics on Quantrex Academy →

More from Atomic Physics