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

Question

The nuclear charge (Ze) is non-uniformly distributed within a nucleus of radius R. The charge density $ρ (r)$ [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. [2008]

Q. For $a = 0$ , the value of $d$ (maximum value of $ρ$ as shown in the figure) is

Smart Achievers · Accepted Answer

(2)

For $a = 0$ , the equation for the graph line is $r = d - \frac{r}{R} d$

Charge in the shaded element $d q = r (4 π r^{2} d r)$

$∴$ $d q = (d - \frac{d}{R} r) 4 π r^{2} d r$

or, $Z e = \int_{0}^{R} 4 π d r^{2} d r - \int_{0}^{R} \frac{4 π d}{R} r^{3} d r$

$\Rightarrow Z e = 4 π d \frac{R^{3}}{3} - \frac{4 π d}{R} \frac{R^{4}}{4}$

$∴$ $\frac{Z e}{4 π d R^{3}} = \frac{1}{3} - \frac{1}{4} = \frac{1}{12}$

$\Rightarrow d = \frac{3 Z e}{π R^{3}}$

Solution

1 $\frac{3 Z e}{4 π R^{3}}$

2 $\frac{3 Z e}{π R^{3}}$

3 $\frac{4 Z e}{3 π R^{3}}$

4 $\frac{Z e}{3 π R^{3}}$

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Solution

1 3Ze4πR3

2 3ZeπR3

3 4Ze3πR3

4 Ze3πR3

Ans. (2) For a=0, the equation for the graph line is r=d-rRd Charge in the shaded element dq=r(4πr2dr) ∴ dq=(d-dRr)4πr2dr or, Ze=∫0R4πd r2dr-∫0R4πdRr3dr ⇒ Ze=4πdR33-4πdRR44 ∴ Ze4πdR3=13-14=112 ⇒ d=3ZeπR3

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1 $\frac{3 Z e}{4 π R^{3}}$

2 $\frac{3 Z e}{π R^{3}}$

3 $\frac{4 Z e}{3 π R^{3}}$

4 $\frac{Z e}{3 π R^{3}}$