Calculating Shortest X-ray Wavelength
The shortest wavelength ($\lambda_{min}$) of X-rays produced is determined by the maximum energy of the incident electrons, which corresponds to the accelerating potential applied. This relationship is described by the Duane-Hunt law.
Duane-Hunt Law Principle
The maximum energy gained by an electron accelerated through a potential $V$ is converted into a single photon of maximum frequency (minimum wavelength). The energy conservation equation is:
$ E_{electron} = E_{photon} $
$ eV = \frac{hc}{\lambda_{min}} $
Where:
- $e$ = elementary charge ($1.602 \times 10^{-19}$ C)
- $V$ = accelerating potential ($50 \text{ KV} = 50 \times 10^3 \text{ V}$)
- $h$ = Planck's constant ($6.626 \times 10^{-34} \text{ J}\cdot\text{s}$)
- $c$ = speed of light ($3 \times 10^8 \text{ m/s}$)
- $\lambda_{min}$ = shortest wavelength
Calculation Steps
- Rearrange the formula to solve for $\lambda_{min}$:
$ \lambda_{min} = \frac{hc}{eV} $
- Use the convenient form $hc \approx 12400 \text{ eV} \cdot \text{\AA}$:
$ \lambda_{min} = \frac{12400 \text{ eV} \cdot \text{\AA}}{V (\text{in Volts})} $
Note: When $V$ is in kilovolts (KV), the formula becomes $\lambda_{min} = \frac{12.4 \text{ keV} \cdot \text{\AA}}{V (\text{in kV})}$.
- Substitute the given accelerating potential ($V = 50$ KV):
$ \lambda_{min} = \frac{12.4 \text{ keV} \cdot \text{\AA}}{50 \text{ kV}} $
- Calculate the minimum wavelength:
$ \lambda_{min} = 0.248 \text{\AA} $
- Rounding the result gives $0.25 \text{\AA}$.
Conclusion
The shortest wavelength present in the X-rays is approximately $0.25 \text{\AA}$. This corresponds to Option C.
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