Q133 / 180 · JIPMER 2015 · Physics
The de-Broglie wavelength of electron falling on the target in an X-ray tube is $\lambda$. The cut-off wavelength of the emitted X-ray is
- A$\lambda_0=\frac{(m c \lambda)^2}{h}$
- B$\lambda_0=\frac{m^2 c \lambda}{h^2}$
- Correct answerC$\lambda_0=\frac{2 m c \lambda^2}{h}$
- D$\lambda_0=\frac{m c \lambda^2}{h^2}$
Solution
The de-Broglie wavelength is given by
$\lambda=\frac{h}{p}=\frac{h}{\sqrt{2 m E}}$ ...(i)
where, $E$ is the energy of the electron. The cut-off wavelength $\lambda_0$ is given by
$\lambda_0=\frac{h C}{E}$ ...(ii)
From Eq. (i),
$\lambda^2=\frac{h^2}{2 m E}$
$\Rightarrow \quad E=\frac{h^2}{2 m X^2}$ ... (iii)
Substituting the value of $E$ from Eq. (ii), we get
$\lambda_0=\frac{h c}{\frac{h^2}{2 m \lambda^2}}=\frac{2 m c \lambda^2}{h}$
Hence, the correct answer is option: 'C'.