The energy of an electron present in Bohr’s second orbit of hydrogen atom is

The energy of an electron present in Bohr’s second orbit of hydrogen atom is

A
$- 1312 J {atom}^{- 1}$
B
$- 328 J {atom}^{- 1}$
C
$- 312 J {atom}^{- 1}$
D
$- 164 J {atom}^{- 1}$

Solution:

The energy expression for the hydrogen atom in $n^{th}$ is

$E_{n} = \frac{- 1312 \times Z^{2}}{n^{2}} kJ / mol \to eqn (1)$

Consider Bohr’s second orbit, n = 2 .

For hydrogen atom, Z =1.

From equation 1, energy of an electron in Bohr’s second orbit is

$E_{2} = \frac{- 1312 \times 1^{2}}{{(2)}^{2}} kJ / mol$

$∴ E_{2} = - 328 kJ / mol$

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