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JEE Advanced levelSingle correct optionDifficulty 3 of 3as090

When a hydrogen-like ion drops from $n = 2$ to $n = 1$ it emits a photon of $91.8 eV$ . Using $E_{1} = - 13.6 eV$ for hydrogen, the ionisation energy of this ion in its ground state is

A $122.4 eV$
B $30.6 eV$
C $54.4 eV$
D $183.6 eV$

Answer: A

91.8 = 13.6 Z^{2} (1 - 1/4) = 10.2 Z^{2}

, so

Z^{2} = 9

,

Z = 3

(

Li X^{2 +}

). Ground-state ionisation energy

= 13.6 \times 9 = 122.4 eV

.

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