Question: Q. 9. Calculate the de-Broglie wavelength of the electron orbiting in the $n=2$ state of hydrogen atom.
U] [Centre 2016]
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Solution:
Ans. Kinetic Energy for the second state
$$ \begin{aligned} E_{k} & =\frac{13.6 \mathrm{eV}}{n^{2}}=\frac{13.6 \mathrm{eV}}{4} \ & =3.4 \times 1.6 \times 10^{-19} \mathrm{~J} \end{aligned} $$
De Broglie’s wavelength,
$$ \begin{aligned} & \lambda=\frac{h}{\sqrt{2 m E_{k}}} \ &=\frac{16.63 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 3.4 \times 1.6 \times 10^{-19}}} \ &=0,067 \mathrm{~nm} \ & \text { [CBSE Marking Scheme 2016] } \end{aligned} $$
