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JEE Main 20199 AprilEvening Shift

Exam: JEE Main
Year: 2019
Date: 9 April
Shift: Evening Shift
Subject: Physics
Chapter: Atomic Physics

JEE Main 2019 · 9 April · Evening Shift · Physics · Atomic Physics

A H e ^ + ion is in its first excited state. Its ionization energy...

A

H{e}^{+}

ion is in its first excited state. Its ionization energy is:

Options

A $13.60eV$
B $48.36eV$
C $54.40eV$
D $6.04eV$

Correct answer

A. $13.60eV$

Step-by-step solution

E = \dfrac{- \left(13.6\right){z}^{2}}{{n}^{2}}

= \dfrac{- \left(13.6\right){\left(2\right)}^{2}}{{2}^{2}} = - 13.60eV

Total energy of

H{e}^{+}

ion is in its first excited state is

- 13.60eV

. Hence, its ionization energy will be

13.60eV

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