QJEEM24Ph0fda: JEE Main 2024 Physics - Class 12 JEEM24Ph0fda Solution: In an alpha particle scattering experiment distance of closest…

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.

($$\frac{1}{4\pi {ϵ}_0}=9\times {10}^9\mathrm{SI}$$ unit, mass of $$\alpha$$ particle $$=6.72\times {10}^{-27}\mathrm{kg}$$)