In an alpha particle scattering experiment distance of closest approach for the particle is . If target nucleus has atomic number 80, then maximum velocity of -particle is ________ m/s approximately.

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Solution

Q.
In an alpha particle scattering experiment distance of closest approach for the $α$ particle is $4.5 \times 10^{- 14} m$ . If target nucleus has atomic number 80, then maximum velocity of $α$ -particle is ________ $\times 10^{5}$ m/s approximately.

$(\frac{1}{4 π ε_{0}} = 9 \times 10^{9}$ SI unit, mass of $α$ particle = $6.72 \times 10^{- 27}$ kg $)$ [2024]

Ans.
(156) $\frac{K Q q}{r_{\min}} = \frac{1}{2} m v^{2} \Rightarrow v = \sqrt{\frac{2 K (Z e) \cdot (2 e) q}{m r_{\min}}} = \sqrt{\frac{4 K Z e^{2}}{m r_{\min}}}$

$\Rightarrow v = \sqrt{\frac{4 \times 9 \times 10^{9} \times 80 \times {(1.6 \times 10^{- 19})}^{2}}{6.72 \times 10^{- 27} \times 4.5 \times 10^{- 14}}}$

$= 1.56 \times 10^{7} m / s = 156 \times 10^{5} m / s$

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