# An alpha particle is accelerated through a potential difference of 100 V. Calculate: (i) The speed acquired by the alpha particle, and

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An alpha particle is accelerated through a potential difference of 100 V. Calculate:

(i) The speed acquired by the alpha particle, and

(ii) The de-Broglie wavelength associated with it. (Take mass of alpha particle = 6.4 x 10-27 kg)

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(i) Kinetic energy of alpha particle K.E. = qv

$\frac12mv^2= 2e \times100$

$\frac12mv^2= 200 \,ev$

$v^2 = \frac{400\,ev}{m}$

$v= \sqrt{\frac{400\,ev}{m}}$

$v = \sqrt{\frac{400\times1.6\times10^{19}}{6.4\times10^{27}}}$

$v =\sqrt{\frac{640}{64}\times10^8}$

$v = \sqrt{100\times10^8}$

v = 10 x 104 m/s

(ii) Kinetic energy gained by alpha particle by potential difference 100 V is

$\frac{p^2}{2m_\alpha}= q_\alpha v$

$\frac{p^2}{2m_\alpha}= 2\,ev$

$p^2 = 4m_\alpha ev$

$p = \sqrt{4m_\alpha ev}$

$p= \sqrt{4\times6.4\times10^{-27}\times1.6\times10^{-19}\times100}$

$p = \sqrt{4096\times10^{-46}}$

$p = 64 \times 10^{-23}$

de-Broglie wavelength,

$\lambda=\frac hp$

$\lambda=\frac{6.64\times10^{-34}}{64\times10^{-23}}$

$\lambda = 0.103 \times 10^{-11}$

$\lambda = 0.0103 \times 10^{-10}$

$\lambda = 0.0103 \,A^o$