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Atoms and Nuclei question

2023 · 11 Apr · Shift 2 · Q48

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Atoms and Nuclei
2023 · 11 Apr · Shift 2 · Q48

Atoms and Nuclei question

2023 · 11 Apr · Shift 2 · Q48

JEE MainPhysicsAtoms and NucleiMCQ+4 / −1

The energy of

\mathrm{He}^{+}

ion in its first excited state is, (The ground state energy for the Hydrogen atom is

-13.6 ~\mathrm{eV})

:

A
$-13.6 ~\mathrm{eV}$
B
$-27.2 ~\mathrm{eV}$
C
$-3.4 ~\mathrm{eV}$
D
$-54.4 ~\mathrm{eV}$

View written solutionFree

Correct answer: A

Use the energy formula for hydrogen-like ions

For a hydrogen-like atom/ion,

E_n = -13.6\,\frac{Z^2}{n^2}\;\text{eV}

where:

$Z$ = atomic number
$n$ = principal quantum number

Identify $Z$ for $\mathrm{He}^+$

Helium has atomic number:

Z=2

Since $\mathrm{He}^+$ is a hydrogen-like ion (only one electron), we can apply the formula directly.

First excited state

The ground state is $n=1$ , so the first excited state is:

n=2

Calculate the energy

E_2 = -13.6\times \frac{2^2}{2^2}

E_2 = -13.6\times \frac{4}{4} = -13.6\,\text{eV}

Match with the options

Thus, the energy of $\mathrm{He}^+$ in its first excited state is:

-13.6\,\text{eV}

So the correct option is A.

Compare with stored correct answer

Stored correct answer: A

Our derived answer: A

They agree.

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