Q5 / 90 · JEE Main 2019 (10 Jan Shift 2) · Chemistry

The ground state energy of a hydrogen atom is $- 13 .6 eV$ . The energy of second excited state of ${He}^{+}$ ion in $eV$ is:

A
$- 27.2$
Correct answer
B
$- 6.04$
C
$- 3.4$
D
$- 54.4$

Solution

$E = - 13 .6 (\frac{z^{2}}{n^{2}}) = - 13 .6 (\frac{2^{2}}{3^{2}}) = - 6 .04 eV$

(Second exited state $=$ Third shell, $n = 3$ )

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

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