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)

An alpha – particle having kinetic energy 5 MeV falls on a Cu-foil.

1.	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)
Answer» `2.35xx10^(-13)` m `1.67 XX 10^(-14) m` `5.98 xx 10^(-15)` `1.67 xx 10^(-16)` m Solution :`K.E. = (K.Ze.2e)/(r)` `r= (9xx10^9 xx 29xx 2xx1.6 xx 10^(-19)^2)/(5XX 1.6 xx 10^(-19) xx 10^(6))` `r= (9xx10^9 xx 29 xx 2 xx 1.6 xx 10^(-19))/(5xx 10^6)` `= 1.67 xx 10^(-14) m`