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Question

The shortest wavelength present in the X-rays at an accelerating potential of 50 KV is :

This question was previously asked in
CUET PG 2026 Agri-Business Management Question Paper (25-Mar-2026) (Shift 2)
The correct answer is
0.25 $\text{\AA}$

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

  1. Rearrange the formula to solve for $\lambda_{min}$:

    $ \lambda_{min} = \frac{hc}{eV} $

  2. 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 $\lambda_{min} = \frac{12.4 \text{ keV} \cdot \text{\AA}}{V (\text{in kV})}$.

  3. Substitute the given accelerating potential ($V = 50$ KV):

    $ \lambda_{min} = \frac{12.4 \text{ keV} \cdot \text{\AA}}{50 \text{ kV}} $

  4. Calculate the minimum wavelength:

    $ \lambda_{min} = 0.248 \text{\AA} $

  5. 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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