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Chapter 17: Problem 2356

If alpha particle and deuteron move with velocity \(\mathrm{V}\) and $2 \mathrm{v}$, the ratio of their de-Brogle wavelength will be............ deuteron (A) \(1: \sqrt{2}\) (B) \(2: 1\) (C) \(1: 1\) (D) \(\sqrt{2}: 1\)

Short Answer

Expert verified
The ratio of the de Broglie wavelengths of the alpha particle and the deuteron is \(\sqrt{2}:1\).

Step by step solution

01

Find the de Broglie wavelength for the alpha particle

We know that de Broglie wavelength is given by \(\lambda=\frac{h}{mv}\). The mass of an alpha particle (which is a Helium nucleus) is \(4m_u\) (where \(m_u\) is the atomic mass unit), and we are given its velocity as V. So, we get the de Broglie wavelength of alpha particle as: \[\lambda_\alpha=\frac{h}{(4m_u)V}\]
02

Find the de Broglie wavelength for the deuteron

Similarly, for the deuteron (which is a nucleus consisting of one proton and one neutron), its mass will be \(2m_u\), and its given velocity is 2V. Therefore, the de Broglie wavelength of the deuteron would be: \[\lambda_d=\frac{h}{(2m_u)(2V)}\]
03

Calculate the ratio of wavelengths

Now, we need to find the ratio of the de Broglie wavelengths of the alpha particle and deuteron: \[\frac{\lambda_\alpha}{\lambda_d}=\frac{\frac{h}{(4m_u)V}}{\frac{h}{(2m_u)(2V)}}\] _multiplying both numerator and denominator by \((4m_u)V(2m_u)(2V)\)_ \[\frac{\lambda_\alpha}{\lambda_d}=\frac{h(2m_u)(2V)}{(4m_u)V}\]
04

Simplify the ratio

By cancelling the terms, we get: \[\frac{\lambda_\alpha}{\lambda_d}=\frac{2}{\sqrt{2}}\] This can be simplified: \[\frac{\lambda_\alpha}{\lambda_d}=\sqrt{2}:1\] So, the correct answer is (D) \(\sqrt{2}:1\).

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