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
B
13.6 eV
C
27.2 eV
D
54.4 eV

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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