Question:
Consider an electron in a hydrogen atom, revolving in its second excited state (having radius $4.65 A$ ). The de-Broglie wavelength of this electron is :
Correct Option: 4,
Solution:
(4) $v=\frac{c}{137 n}=\frac{c}{137 \times 3}$
$\lambda=\frac{h}{p}=\frac{h}{m v}=\frac{h}{\left(\frac{m \times c}{3 \times 137}\right)}=\frac{h}{m c} \times(3 \times 137)=9.7 A$
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