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. `( 3 Ze)/(4 pi R^(3))`
B. `(3Ze)/(pi R^(3))`
C. `( 4Z e)/( 3pi R^(3))`
D. `(Ze)/(3pi R^(3))`

Answer

B

Explanation

For `a=0rArrp(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))`