# An α - Particle Is Accelerated Through A Potential Difference Of V Volts From Rest. The De-Broglie Wavelength Associated With It Is (In Angstrom):

Kinetic Energy gained by the α-particle by potential difference V is,

$\frac{P^{2}}{2m_{\alpha}} = q_{\alpha}V = 2eV \\ \Rightarrow m_{\alpha} = 4m_{p}\\ \Rightarrow p^{2} = 16m_{p}eV\\ \Rightarrow p = 4\sqrt{m_{p}eV}\\ \Rightarrow 4\sqrt{(1.67 * 10^{-27})kg * (1.6 * 106 – 19) C * V}\\ \Rightarrow 4 * 1.634 * 10^{-23} \sqrt{V}\\ \lambda =\frac{h}{p} = \frac{6.62 * 10^{-34}}{4 * 1.634 * 10^{-23}\sqrt{V}}\\ \Rightarrow \frac{0.101^{\circ}}{\sqrt{V}}A$

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