If the potential difference applied to tube is doubled and the separation between the filament and the target is also doubled, the cutoff wavelength

Options

A. will remain unchanged
B. will be doubled
C. will be halved
D. will become four times the original

Answer

To solve the problem regarding the effect of doubling the potential difference and the separation between the filament and the target on the cutoff wavelength, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship**: The cutoff wavelength (λ_min) is given by the equation:
\[ \lambda_{\text{min}} = \frac{hc}{eV} \]
where:
- \( h \) is Planck's constant,
- \( c \) is the speed of light,
- \( e \) is the charge of an electron,
- \( V \) is the potential difference.

2. **Identify the changes**: According to the problem, the potential difference \( V \) is doubled (i.e., \( V \) becomes \( 2V \)), and the separation between the filament and the target is also doubled. However, the separation does not affect the cutoff wavelength.

3. **Substitute the new potential difference**: With the new potential difference, we can express the new cutoff wavelength:
\[ \lambda'_{\text{min}} = \frac{hc}{e(2V)} = \frac{hc}{2eV} \]

4. **Relate the new cutoff wavelength to the original**: We can relate the new cutoff wavelength to the original cutoff wavelength:
\[ \lambda'_{\text{min}} = \frac{1}{2} \lambda_{\text{min}} \]
This shows that the new cutoff wavelength is half of the original cutoff wavelength.

5. **Conclusion**: Since the cutoff wavelength is inversely proportional to the potential difference, doubling the potential difference results in halving the cutoff wavelength. Therefore, the correct answer is that the cutoff wavelength is halved.

### Final Answer:
The cutoff wavelength will be halved (Option 3).

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