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# If stationary proton and $\;\alpha-\;$ particle dre accelerated through same potential difference , the ratio of Debrogle's wavelength will be

$(a)\;2\qquad(b)\;1\qquad(c)\;2 \sqrt{2}\qquad(d)\;None$

Answer : (c) $\;2 \sqrt{2}$
Explanation :
The gain in K.E of a charge particle of the moving through a potential difference of V is given as eV ,that is also equal to $\;\large\frac{1}{2}\;mV^2\;$ where V is velocity of charge particle . Disregarding relatiuistic effect
$\large\frac{1}{2}\;mV^2=qV \to V=\sqrt{\large\frac{2qV}{m}}$
$mV=\sqrt{2mqV}$
De-Brogli's wavelength
$=\lambda=\large\frac{h}{mV}=\large\frac{h}{\sqrt{2mqV}}$
$\large\frac{\lambda_{P}}{\lambda_{\alpha}}=\sqrt{\large\frac{m_{\alpha}q_{\alpha}V_{\alpha}}{m_{P}q_{P}V_{p}}}$
Puting $V_{\alpha}=V_{P}\;,\large\frac{\lambda_{P}}{\lambda_{\alpha}}=\sqrt{4\times2}=2 \sqrt{2}\;.$