Question:
An electron accelerated through a potential difference \(V_1\) has a de-Broglie wavelength of \(λ\). When the potential is changed to \(V_2\), its de-Broglie wavelength increases to \(2λ\). The value of \((V_1/V_2)\) is equal to
An electron accelerated through a potential difference \(V_1\) has a de-Broglie wavelength of \(λ\). When the potential is changed to \(V_2\), its de-Broglie wavelength increases to \(2λ\). The value of \((V_1/V_2)\) is equal to
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For an electron, \(\lambda\propto\frac{1}{\sqrt V}\).
Updated On: Oct 1, 2026
- \(3:1\)
- \(9:4\)
- \(3:2\)
- \(4:1\)
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept:
An electron accelerated through \(V\) has \(\lambda = \frac{h}{\sqrt{2meV}}\), so \(\lambda\propto V^{-1/2}\).
Step 2: Compare:
\(\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{V_1}{V_2}} = 2\), so \(\frac{V_1}{V_2} = 4\).
Step 3: Check:
Doubling the wavelength means quartering the voltage, which fits \(V_1:V_2 = 4:1\).
Final Answer:
The ratio \(V_1:V_2\) is \(4:1\), option (D). \[ \boxed{4:1} \]
An electron accelerated through \(V\) has \(\lambda = \frac{h}{\sqrt{2meV}}\), so \(\lambda\propto V^{-1/2}\).
Step 2: Compare:
\(\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{V_1}{V_2}} = 2\), so \(\frac{V_1}{V_2} = 4\).
Step 3: Check:
Doubling the wavelength means quartering the voltage, which fits \(V_1:V_2 = 4:1\).
Final Answer:
The ratio \(V_1:V_2\) is \(4:1\), option (D). \[ \boxed{4:1} \]
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