The energy of a hydrogen atom in the ground state is -13.6 eV . The energy of a He^ + ion in the first excited state will be

Correct answer: A. -13.6 eV Energy E of an atom with principal quantum number n is given by E= -13.6 n^2 Z^2 for first excited state n=2 and for He^ + Z=2 aligned & =(-136 ×(2)^2)/((2)^2) & =-13.6 eV aligned

NEET 2010

Exam: NEET
Year: 2010
Subject: Physics
Chapter: Atomic Physics

NEET 2010 · Physics · Atomic Physics

The energy of a hydrogen atom in the ground state is -13.6 eV . The...

The energy of a hydrogen atom in the ground state is

-13.6 \mathrm{eV}

. The energy of a

\mathrm{He}^{+}

ion in the first excited state will be

Options

A $-13.6 \mathrm{eV}$
B $-27.2 \mathrm{eV}$
C $-54.4 \mathrm{eV}$
D $-6.8 \mathrm{eV}$

Correct answer

A. $-13.6 \mathrm{eV}$

Step-by-step solution

Energy E of an atom with principal quantum number

\mathrm{n}

is given by

E=\frac{-13.6}{\mathrm{n}^2} Z^2

for first excited state

\mathrm{n}=2

and for

\mathrm{He}^{+} \mathrm{Z}=2

\begin{aligned} \Rightarrow \quad\mathrm{E}& =\frac{-136 \times(2)^2}{(2)^2}\\ & =-13.6 \mathrm{eV} \end{aligned}

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