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12.Atoms
hard
In an alpha particle scattering experiment distance of closest approach for the $\alpha$ particle is $4.5 \times 10^{-14} \mathrm{~m}$. If target nucleus has atomic number $80$ , then maximum velocity of $\alpha$-particle is . . . . .. $\times 10^5$ $\mathrm{m} / \mathrm{s}$ approximately.
$\left(\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \mathrm{SI}\right.$ unit, mass of $\alpha$ particle $=$ $\left.6.72 \times 10^{-27} \mathrm{~kg}\right)$
A
$155$
B
$156$
C
$157$
D
$158$
(JEE MAIN-2024)
Solution
$v=\sqrt{\frac{4 \mathrm{KZe}^2}{\mathrm{mr}_{\min }}}$
$=\sqrt{\frac{4 \times 9 \times 10^9 \times 80}{6.72 \times 10^{-27} \times 4.5 \times 10^{-14}}} \times 1.6 \times 10^{-19}$
$=9.759 \times 10^{25} \times 1.6 \times 10^{-19}$
$=156 \times 10^5 \mathrm{~m} / \mathrm{s}$
Standard 12
Physics
