Q187 / 200 · WBJEE 2014 · Physics
The de-Broglie wavelength of an electron is the same as that of a $50 \mathrm{keV}$ X-ray photon. The ratio of the energy of the photon to the kinetic energy of the electron is (the energy equivalent of electron mass is $0.5 \mathrm{MeV}$ )
- A01:50
- B01:20
- Correct answerC20: 1
- D50: 1
Solution
de-Broglie wavelength $\lambda=\frac{h}{\sqrt{2 m K}}$
The kinetic energy of the electron
$$
K_{\text {electron }}=\frac{1}{2 m} \cdot \frac{h^{2}}{\lambda^{2}}
$$
Where $h=$ Planck constant $\lambda=$ wavelength
The photon energy
$$
E_{\text {phocton }}=\frac{h c}{\lambda}
$$
From Eqs. (ii) and (i), we get
$\frac{E_{\text {photon }}}{K_{\text {electron }}}=\frac{h c / \lambda}{h^{2} / 2 m \cdot \lambda^{2}}=\frac{h c \cdot \lambda^{2} \times 2 m}{h^{2} \cdot \lambda}$
where $m=0.5 \mathrm{MeV}=5 \times 10^{5} \mathrm{eV}$
$\begin{aligned} \frac{h}{\lambda c}=50 \times 10^{3} \mathrm{eV}=\frac{2 \mathrm{m} \lambda c}{h} \\=\frac{2 \mathrm{m}}{h / \lambda c} &=\frac{2 \times 5 \times 10^{5}}{50 \times 10^{3}} \quad\left(\because \mathrm{m}=\frac{h}{c \lambda}\right) \\ &=20: 1 \end{aligned}$
Hence, the correct answer is option: 'C'.