An α -particle having energy 10MeV collides with a nucleus of atomic number 50. Then distance of closest approach will be

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

14.4×10−16m

1.7×10−7m

1.5×10−12m

14.4×10−15m

answer is D.

(Unlock A.I Detailed Solution for FREE)

Detailed Solution

KE=10MeV=10×106ev; Z=50; q1=2e, q2=50eKE=14π∈0.q1q2d∴d=9×109×2e×50e10×106e =9×109×100×16×10−19107 =14.4×10−15m

Watch 3-min video & get full concept clarity