Radius of a nucleus is given by the relation `R = R_(0)A^(1//3)` where `R_(0) = 1.3 xx 10^(-15) m` and A is mass number. For a nucleon inside a nucleus, de-Broglie wavelength is given by diameter of the nucleus. Average kinetic energy of a nucleon in the `Te^(128)` nucleus based on above information will be :

Options

A. 4.7 MeV
B. 10 MeV
C. 2 MeV
D. 12 MeV

Explanation

`lambda=2R=2R_(0)A^(1//3)`
`P=h/(lambda)implies E=(P^(2))/(2m)=(h^(2))/(2m(4R_(0)^(2)A^(2//3)))`
`E=((6.62xx10^(-34))^(2))/(2xx1.67xx10^(-27)xx4(1.3xx10^(-15))^(2)(128)^(2//3))` joule
=4.72 MeV.