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 $=$ $\left.6.72 \times 10^{-27} \mathrm{~kg}\right)$
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 $=$ $\left.6.72 \times 10^{-27} \mathrm{~kg}\right)$
- [JEE MAIN 2024]
- A
$155$
- B
$156$
- C
$157$
- D
$158$
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