A proton moving with one tenth of velocity of light has a certain de Broglie wavelength of $\lambda$. An alpha particle having certain kinetic energy has the same de-Brogle wavelength $\lambda$. The ratio of kinetic energy of proton and that of alpha particle is:
Solution
For same $\lambda_{1}$ momentum should be same,
<br/><br/>$(P)_{P}=(P)_{\alpha}$
<br/><br/>$\Rightarrow \sqrt{2 k_{P} m_{P}}=\sqrt{2 k_{\alpha} m_{\alpha}}$
<br/><br/>$\Rightarrow k_{P} m_{P}=k_{\alpha} m_{\alpha}$
<br/><br/>$\Rightarrow \frac{k_{P}}{k_{\alpha}}=\left(\frac{m_{\alpha}}{m_{P}}\right)=\frac{4}{1}=4: 1$
About this question
Subject: Physics · Chapter: Dual Nature of Matter and Radiation · Topic: de Broglie Hypothesis
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