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

Answer

B

Explanation

KE of `alpha` particales= repulsive force at distance .r. =- energy of `alpha`- particle
`K.E. =(K.Ze. 2e)/(r ) , 5xx10^(6) xx1.6xx10^(-19) J =(9xx10^(9) xx29xx2xx(1.6xx10^(-19))^(2))/(r )`
`r=(9xx10^(9) xx29xx2xx(1.6xx10^(-19))^(2))/(5xx1.6xx10^(-19)xx10^(6))=(9xx10^(9) xx29xx2xx1.6xx10^(-19))/(5xx10^(6))=1.67xx10^(-14) m`