Question

If electrons of mass 'm' and de Broglie wavelength '$\lambda$', incident on a metal plate in an X-ray tube, then the minimum wavelength of X-rays emitted is
(h = Planck's constant and c = Speed of light in vacuum)

• A. $\frac{4mc\lambda^2}{h}$
• B. $\frac{2mc\lambda^2}{h}$
• C. $\frac{mc\lambda^2}{4h}$
• D. $\frac{mc\lambda^2}{2h}$

Hint:

Relate wave properties: Particle $\lambda \rightarrow$ Momentum $\rightarrow$ Energy $\rightarrow$ Photon $\lambda_{min}$.

Answer 1

The Correct Option is B

Step 1: Kinetic Energy from de Broglie Wavelength:
The de Broglie wavelength of an electron is given by $\lambda = \frac{h}{p}$, where $p$ is momentum. So, momentum $p = \frac{h}{\lambda}$. The Kinetic Energy ($K$) of the electron is: \[ K = \frac{p^2}{2m} = \frac{(h/\lambda)^2}{2m} = \frac{h^2}{2m\lambda^2} \]
Step 2: X-ray Cut-off Wavelength:
The entire kinetic energy of the electron is converted into a single photon for the minimum wavelength ($\lambda_{min}$) of X-rays produced (Duane-Hunt Law). \[ K = \frac{hc}{\lambda_{min}} \]
Step 3: Equating and Solving:
\[ \frac{h^2}{2m\lambda^2} = \frac{hc}{\lambda_{min}} \] Solving for $\lambda_{min}$: \[ \lambda_{min} = \frac{hc \cdot 2m\lambda^2}{h^2} \] \[ \lambda_{min} = \frac{2mc\lambda^2}{h} \]
Step 4: Final Answer:
The minimum wavelength is $\frac{2mc\lambda^2}{h}$.