JEE MAIN - Physics (2023 - 1st February Morning Shift - No. 11)
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:
1 : 4
2 : 1
4 : 1
1 : 2
Explanation
For same $\lambda_{1}$ momentum should be same,
$(P)_{P}=(P)_{\alpha}$
$\Rightarrow \sqrt{2 k_{P} m_{P}}=\sqrt{2 k_{\alpha} m_{\alpha}}$
$\Rightarrow k_{P} m_{P}=k_{\alpha} m_{\alpha}$
$\Rightarrow \frac{k_{P}}{k_{\alpha}}=\left(\frac{m_{\alpha}}{m_{P}}\right)=\frac{4}{1}=4: 1$
$(P)_{P}=(P)_{\alpha}$
$\Rightarrow \sqrt{2 k_{P} m_{P}}=\sqrt{2 k_{\alpha} m_{\alpha}}$
$\Rightarrow k_{P} m_{P}=k_{\alpha} m_{\alpha}$
$\Rightarrow \frac{k_{P}}{k_{\alpha}}=\left(\frac{m_{\alpha}}{m_{P}}\right)=\frac{4}{1}=4: 1$
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