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Question: The energy of a proton and an\(\alpha\)particle is the same. Then the ratio of the de-Broglie wavele...

The energy of a proton and anα\alphaparticle is the same. Then the ratio of the de-Broglie wavelengths of the proton and the α\alphais

A

1 : 2

B

2 : 1

C

1 : 4

D

4 : 1

Answer

2 : 1

Explanation

Solution

By using λ=h2mE\lambda = \frac{h}{\sqrt{2mE}} ⇒ λ∝1m\lambda \propto \frac{1}{\sqrt{m}} (E – same)

⇒ λprotonλα−particle=mαmp=21\frac{\lambda_{proton}}{\lambda_{\alpha - particle}} = \sqrt{\frac{m_{\alpha}}{m_{p}}} = \frac{2}{1}