If and `alpha`-paricle of mass m, charged q and velocity v is incident on a nucleus charge Q and mass m, then the distance of closest approach is

Question

If and `alpha`-paricle of mass m, charged q and velocity v is incident on a nucleus charge Q and mass m, then the distance of closest approach is

If and `alpha`-paricle of mass m, charged q and velocity v is incident on a nucleus charge Q and mass m, then the distance of closest approach is
A. `(Qq)/(4pi epsi_(0)m^(2))`
B. `(Qq)/(2pi epsi_(0)mv^(2))`
C. `(Qq mv^(2))/(2)`
D. `(Qq)/(mv^(2))`

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - B
According to the question,
For the distance of closest approach kinetic energy will totally be converted to potential energy
Hence, `" " (1)/(2) mv^(2) =(1)/(4piepsi_(0)) (Qq)/(r_(0)) rArr r_(0)=(Qq)/(2piepsi_(0)mv^(2))`

4 views Answered 2023-02-05 15:29