The nuclear charge (`Ze`) is non uniformlly distribute with in a nucleus of radius `r`. The charge density `rho(r)` (charge per unit volume) is dependent only on the radial distance r form the centre of the nucleus s shown in figure. The electric field is only along the radial direction.

For a=0 the value of d (maximum value of `rho` as shown in the figure) is

Options

A. `(3Ze)/(4piR^3)`
B. `(3Ze)/(piR^3)`
C. `(4Ze)/(3piR^3)`
D. `(Ze)/(3piR^3)`

Answer

B

Explanation

For `a=0`,
`rho(r)=(-d/R.r+d)`
Now, `int_0^R(4pir^2)(d-d/Rr)dr`=net charge =Ze
Solving this equation we get `d=(3Ze)/(piR^3)`