Question

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. \(\frac{9}{4}\)
• B. \(\frac{4}{9}\)
• C. \(\frac{2}{3}\)
• D. \(\frac{3}{2}\)

Hint:

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.

Answer 1

The Correct Option is A

The de-Broglie wavelength \(\lambda\) of a particle is given by the equation: \[ \lambda = \frac{h}{p} \] where \(h\) is Planck's constant and \(p\) is the momentum of the particle. The momentum \(p\) of a particle accelerated by a potential difference \(V\) is given by: \[ p = \sqrt{2m e V} \] where \(m\) is the mass of the particle, \(e\) is the charge of the particle, and \(V\) is the potential difference. Now, we have two situations: 1. For the first potential difference \(V_1\), the de-Broglie wavelength is \(\lambda_1\), and the momentum is \(p_1 = \sqrt{2m e V_1}\). 2. For the second potential difference \(V_2\), the de-Broglie wavelength is \(\lambda_2 = \frac{3\lambda_1}{2}\), and the momentum is \(p_2 = \sqrt{2m e V_2}\). The relation between the wavelengths and the momenta is: \[ \frac{\lambda_2}{\lambda_1} = \frac{p_1}{p_2} = \frac{\sqrt{2m e V_1}}{\sqrt{2m e V_2}} \] Substituting \(\lambda_2 = \frac{3\lambda_1}{2}\) into the equation: \[ \frac{3}{2} = \frac{\sqrt{V_1}}{\sqrt{V_2}} \] Squaring both sides: \[ \frac{9}{4} = \frac{V_1}{V_2} \] Thus, the ratio \(\frac{V_1}{V_2} = \frac{9}{4}\). Final Answer: Option (A) \(\frac{9}{4}\).