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Question

If the energy of hydrogen atom in the ground state is $$-13.6\ eV$$, then energy of $$He^{+}$$ ion in first excited state will be


A
6.8 eV
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B
13.6 eV
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C
27.2 eV
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D
54.4 eV
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Solution

The correct option is B $$-13.6\ eV$$
$$E_{He} = -\dfrac {Z^{2}}{n^{2}} \times 13.6\ eV$$
For $$He^{+}$$ ion $$Z = 2$$ and for first excited state $$n = 2$$
$$\therefore E_{He^{+}} = -\dfrac {2^{2}}{2^{2}}\times 13.6\ eV$$
$$= -13.6\ eV$$.

Physics

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