MCQ
If $\lambda_1$ and $\lambda_2$ denote the wavelengths of de Broglie waves for electrons in the first and second Bohr orbits in a hydrogen atom, then $\lambda_1/\lambda_2$ is equal to
- A$2/1$
- ✓$1/2$
- C$1/4$
- D$4/1$
✓
Answer
Correct option: B.
$1/2$
b
For the incident electron $\frac{1}{2} \mathrm{mv}^{2}=\mathrm{eV}$
For the incident electron $\frac{1}{2} \mathrm{mv}^{2}=\mathrm{eV}$
$\mathrm{or} \mathrm{p}^{2}=2 \mathrm{meV}$
$\therefore $ de-Broglie wavelength $\lambda_{1}=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{meV}}}$
Shortest X-ray wavelength $\lambda_{2}=\frac{\mathrm{hc}}{\mathrm{eV}}$
$\therefore \frac{\lambda_{1}}{\lambda_{2}}=\frac{1}{c} \sqrt{\left(\frac{V}{2}\right)\left(\frac{e}{m}\right)}=\frac{\sqrt{\frac{10^{4}}{2} \times 1.8 \times 10^{11}}}{3 \times 10^{8}}=0.1$
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