Looking for classes? Ksquare Career Institute, Bengaluru →
Energy E of a hydrogen atom with principal quantum number n is given by $E = \dfrac{-13.6}{n^2}$ eV. The energy of a photon ejected when the electron jumps from n = 3 state to n = 2 state of hydrogen is approximately :-
A
0.85 eV
B
3.4 eV
C
1.9 eV
D
1.5 eV
Detailed Solution
Energy of photon $= E_3 - E_2$
$E_3 = \dfrac{-13.6}{9} = -1.51$ eV
$E_2 = \dfrac{-13.6}{4} = -3.4$ eV
$E_3 - E_2 = \left(\dfrac{-13.6}{9}\right) - \left(\dfrac{-13.6}{4}\right) = 13.6\left(\dfrac{1}{4} - \dfrac{1}{9}\right)$
$= \dfrac{5}{36} \times 13.6 \approx 1.9$ eV
This is the first line of the Balmer series ($H_\alpha$, red light of wavelength about 656 nm).
$E_3 = \dfrac{-13.6}{9} = -1.51$ eV
$E_2 = \dfrac{-13.6}{4} = -3.4$ eV
$E_3 - E_2 = \left(\dfrac{-13.6}{9}\right) - \left(\dfrac{-13.6}{4}\right) = 13.6\left(\dfrac{1}{4} - \dfrac{1}{9}\right)$
$= \dfrac{5}{36} \times 13.6 \approx 1.9$ eV
This is the first line of the Balmer series ($H_\alpha$, red light of wavelength about 656 nm).