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Question: What will be the ratio of de-Broglie wavelengths of proton and \(\alpha -\)particle of same energy....

What will be the ratio of de-Broglie wavelengths of proton and α−\alpha -particle of same energy.

A

2 : 1

B

1 : 2

C

4 : 1

D

1 : 4

Answer

2 : 1

Explanation

Solution

λ=h2mE⇒λ∝1m⇒λpλα=mαmp=21\lambda = \frac{h}{\sqrt{2mE}} \Rightarrow \lambda \propto \frac{1}{\sqrt{m}} \Rightarrow \frac{\lambda_{p}}{\lambda_{\alpha}} = \sqrt{\frac{m_{\alpha}}{m_{p}}} = \frac{2}{1}