MCQ
If the potential difference applied to the tube is doubled and the separation between the filament and the target is also doubled, the cutoff wavelength:
  • A
    Will remain unchanged.
  • B
    Will be doubled.
  • C
    Will be halved.
  • D
    Will become four times the original.

Answer

  1. Will be halved.

Explanation:

Cut off wavelength is given by

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

where h = Planck's constan

c = speed of light

e = charge on an electron

V = potential difference applied to the tube

When potential difference (V) applied to the tube is doubled, cutoff wavelength

$\big({\lambda}'_\text{min}\big)$ is given by

${\lambda}'_\text{min}=\frac{\text{hc}}{\text{e}(2\text{V})}$

$\Rightarrow\lambda'_\text{min}=\frac{\lambda_\text{min}}{2}$

Cuttoff wavelength does not depend on the separation between the filament and the target.

Thus, cutoff wavelength will be halved if the the potential difference applied to the tube is doubled.

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