Paragraph: The nuclear charge (Ze) is non-uniformly distributed within a nucleus of radius R . The charge density (r) [charge per unit volume] is dependent only on the radial distance r from the centre of the nucleus as

Correct answer: B. (3 Z e)/(π R^3) For a=0 aligned & (r)=(-(d)/(R) · r+d) & Now ∫_0^R(4 π r^2)(d-(d)/(R) r) d r= net charge =Z e aligned Solving this equation, we get d= (3 Z e)/(π R^3)

JEE Advanced 2008Paper 2

Exam: JEE Advanced
Year: 2008
Paper: Paper 2
Subject: Physics
Chapter: Electrostatics

JEE Advanced 2008 · Paper 2 · Physics · Electrostatics

Paragraph: The nuclear charge (Ze) is non-uniformly distributed...

Paragraph: The nuclear charge

(\mathrm{Ze})

is non-uniformly distributed within a nucleus of radius

R

. The charge density

\rho(r)

[charge per unit volume] is dependent only on the radial distance

r

from the centre of the nucleus as shown in figure. The electric field is only along the radial direction. Paragraph: The nuclear charge (Ze) is non-uniformly distributed within a nucleus of radius R . The charge density (r)

Paragraph: The nuclear charge (Ze) is non-uniformly distributed within a nucleus of radius R . The charge density (r)

Question: For

a=0

, the value of

d

(maximum value of

\rho

as shown in the figure) is

Options

A $\frac{3 Z e}{4 \pi R^3}$
B $\frac{3 Z e}{\pi R^3}$
C $\frac{4 Z e}{3 \pi R^3}$
D $\frac{Z e}{3 \pi R^3}$

Correct answer

B. $\frac{3 Z e}{\pi R^3}$

Step-by-step solution

For

a=0

\begin{aligned} & \rho(r)=\left(-\frac{d}{R} \cdot r+d\right) \\ & \text{ Now } \int_0^R\left(4 \pi r^2\right)\left(d-\frac{d}{R} r\right) d r=\text{ net charge }=Z e \end{aligned}

Solving this equation, we get

\frac{3 Z e}{\pi R^3}

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