The ratio of de broglie wavelength of electron accelerated through potentials `V_(1)` and `V_(2)` is 1:2 the ratio of potentials `V_(1) : V_(2)` is

Options

A. `4:1`
B. `1:4`
C. `1:2`
D. `2:1`

Answer

To solve the problem, we need to find the ratio of potentials \( V_1 : V_2 \) given that the ratio of the de Broglie wavelengths of electrons accelerated through these potentials is \( 1 : 2 \).

### Step-by-Step Solution:

1. **Write the Given Ratio of Wavelengths:**
\[ \frac{\lambda_1}{\lambda_2} = \frac{1}{2} \]

2. **Use the Formula for de Broglie Wavelength:**
The de Broglie wavelength \( \lambda \) of an electron can be expressed as:
\[ \lambda = \frac{h}{\sqrt{2m \cdot KE}} \]
where \( KE \) (kinetic energy) for an electron accelerated through a potential \( V \) is given by:
\[ KE = eV \]
Therefore, we can rewrite the de Broglie wavelength as:
\[ \lambda = \frac{h}{\sqrt{2m \cdot eV}} \]

3. **Express Wavelengths for Both Potentials:**
For potentials \( V_1 \) and \( V_2 \):
\[ \lambda_1 = \frac{h}{\sqrt{2m \cdot eV_1}} \quad \text{and} \quad \lambda_2 = \frac{h}{\sqrt{2m \cdot eV_2}} \]

4. **Set Up the Ratio of Wavelengths:**
Substitute the expressions for \( \lambda_1 \) and \( \lambda_2 \) into the ratio:
\[ \frac{\lambda_1}{\lambda_2} = \frac{\frac{h}{\sqrt{2m \cdot eV_1}}}{\frac{h}{\sqrt{2m \cdot eV_2}}} = \frac{\sqrt{2m \cdot eV_2}}{\sqrt{2m \cdot eV_1}} = \sqrt{\frac{V_2}{V_1}} \]

5. **Equate the Ratios:**
From step 1, we have:
\[ \frac{1}{2} = \sqrt{\frac{V_2}{V_1}} \]

6. **Square Both Sides:**
Squaring both sides gives:
\[ \left(\frac{1}{2}\right)^2 = \frac{V_2}{V_1} \implies \frac{1}{4} = \frac{V_2}{V_1} \]

7. **Find the Ratio of Potentials:**
Rearranging gives:
\[ \frac{V_1}{V_2} = 4 \implies V_1 : V_2 = 4 : 1 \]

### Final Answer:
The ratio of potentials \( V_1 : V_2 \) is \( 4 : 1 \).

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