Looking for classes? Ksquare Career Institute, Bengaluru →
The energy of second Bohr orbit of the hydrogen atom is $-328$ kJ mol$^{-1}$; hence the energy of fourth Bohr orbit would be :-
A
$-1312$ kJ mol$^{-1}$
B
$-82$ kJ mol$^{-1}$
C
$-41$ kJ mol$^{-1}$
D
$-164$ kJ mol$^{-1}$
Detailed Solution
Energy of the $n$th Bohr orbit: $E_n \propto -\dfrac{1}{n^2}$
$\dfrac{E_4}{E_2} = \dfrac{n_2^2}{n_4^2} = \dfrac{2^2}{4^2} = \dfrac{4}{16} = \dfrac{1}{4}$
$E_4 = \dfrac{E_2}{4} = \dfrac{-328}{4}$
$E_4 = -82$ kJ mol$^{-1}$
(Check: $E_1 = -1312$ kJ mol$^{-1}$, so $E_4 = \dfrac{-1312}{16} = -82$ kJ mol$^{-1}$.)
$\dfrac{E_4}{E_2} = \dfrac{n_2^2}{n_4^2} = \dfrac{2^2}{4^2} = \dfrac{4}{16} = \dfrac{1}{4}$
$E_4 = \dfrac{E_2}{4} = \dfrac{-328}{4}$
$E_4 = -82$ kJ mol$^{-1}$
(Check: $E_1 = -1312$ kJ mol$^{-1}$, so $E_4 = \dfrac{-1312}{16} = -82$ kJ mol$^{-1}$.)