Q89 / 90 · JEE Main 2014 (09 Apr Online) · Physics

If the binding energy of the electron in a hydrogen atom is $13.6 eV$ , the energy required to remove the electron from the first excited state of ${Li}^{+ +}$ is :

A
$13.6 eV$
B
$3.4 eV$
C
$122.4 eV$
Correct answer
D
$30.6 eV$

Solution

Energy of an electron in n^th orbit of an ion with atomic number

z

is,

E = - 13 .6 \times \frac{z^{2}}{n^{2}} eV

for

1^{st}

excited state of

L i^{+ +}, z = 3, n = 2

E = - 13 .6 \times \frac{3^{2}}{n^{2}} = - 13 .6 \times \frac{9}{4} eV

= - 30 .6 eV

∴

Required energy

= 30.6 eV

Hence, the correct answer is option: 'D'.

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