Question:medium
A charge particle when accelerated from rest by a potential difference \(V_1\) has a de-Broglie wavelength \(\lambda_1\), and when accelerated by a potential difference \(V_2\) has a de-Broglie wavelength \(\lambda_2\). If \(\lambda_2 = \frac{3\lambda_1}{2}\), find the ratio \(\frac{V_1}{V_2}\).
A charge particle when accelerated from rest by a potential difference \(V_1\) has a de-Broglie wavelength \(\lambda_1\), and when accelerated by a potential difference \(V_2\) has a de-Broglie wavelength \(\lambda_2\). If \(\lambda_2 = \frac{3\lambda_1}{2}\), find the ratio \(\frac{V_1}{V_2}\).
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The de-Broglie wavelength is inversely proportional to the momentum of the particle. The momentum of a charged particle accelerated by a potential difference is proportional to the square root of the potential difference.
Updated On: Apr 7, 2026
- \(\frac{9}{4}\)
- \(\frac{4}{9}\)
- \(\frac{2}{3}\)
- \(\frac{3}{2}\)
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The Correct Option is A
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