An `alpha`-particle having kinetic energy 5 MeV falls on a Cu-foil. The shortest distance from the nucleus of Cu to which `alpha`-particle reaches is (Atomic no. of `Cu = 29, K = 9xx10^(9)" Nm"^(2)//C^(2)`)

Options

A. `2.35 xx 10^(-13)m`
B. `1.67xx10^(-14)m`
C. `5.98 xx 10^(-15)m`
D. `6.32 xx 10^(-15)m`

Answer

B

Explanation

KE of particles = repulsive force at distance .r. = - energy of `alpha` -particle
`K.E.=(K.Ze.2e)/(r ), 5xx10^(6)xx1.6 xx10^(-19)J =(9xx10^(9) xx 29xx2xx(1.6xx10^(-19))^(2))/(r )`
`r=(9xx10^(9) xx 29 xx 2xx(1.6 xx10^(-19))^(2))/(5xx1.6 xx 10^(-19)xx10^(6)) =(9xx10^(9)xx29xx2xx1.6 xx 10^(-19))/(5xx10^(6))=1.67xx10^(-14)m`