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An energetic a-particle experiences head...

An energetic a-particle experiences head-on collision with nucleus having Z = 85. If distance of closest approach is `1.85xx10^(-14)m` then calculate energy of a particle.

A

23.13 MeV

B

13.2 MeV

C

10 MeV

D

20 MeV

Text Solution

Verified by Experts

The correct Answer is:
B
13.2 MeV
Kinetic energy of a particle = P.E. at `r_(0)` distance
`=(kxxZexx2e)/(r_(0)xxe)toeV`
`=(9xx10^(9)xx85kxx2xxe^(2))/(1.85xx10^(-14)xxe)`
`=(9xx10^(9)xx170xx1.6xx10^(-19))/(1.85xx10^(-14))`
`=1323.24xx10^(4)eV`
`=13.23xx10^(6)eV`
`=13.23MeV`
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