1000 Solved Problems in Modern Physics

8.3 Solutions 449

ΔEc=

3

5

1

4 πε 0

e^2

R

[Z(Z+1)−Z(Z−1)]

=

6 Ze^2

5 R

e^2

4 πε 0

= 1. 2 × 1. 44

Z

R

MeV-fm

EquatingΔEctoMSi−MA= 3. 48 + 2 × 0. 51 = 4 .5 MeV, andZ=13,

we findRfrom whichro=R/A^1 /^3 can be determined, whereA=27. Thus

ro= 1 .66 fm.

8.5ΔPx.Δx=(uncertainty principle)

orcΔPx=

c

Δx

=

197 .3MeV−fm

5fm

= 39 .6MeV

The kinetic energyT=

c^2 p^2

2 Mc^2

=

(39.6)^2

2 × 940

= 0 .83 MeV

8.6 The mass-energy equation for positron decay is

M(^14 O)−M(^14 N)= 2 me+

Eβ(max)+Eγ

931. 5

= 2 × 0. 000548 +

1. 835 + 2. 313

931. 5

= 0. 005549

or M(^14 O)= 14. 003074 + 0. 005549 = 14 .008623 amu

8.3.2 ElectricPotentialandEnergy ........................

8.7 The charge densityρ= 3 ze/ 4 πR^3. Consider a spherical shell of radiirand
r+dr.
The volume of the shell is 4πr^2 dr. The charge in the shellq′=(4πr^2 dr)ρ.
The electrostatic energy due to the chargeq′and the charge (q′′) of the sphere
of radiusr, which is^43 πr^3 ρ, is calculated by imaginingq′′to be deposited at
the centre. The charge outside the sphere of radiusrdoes not contribute to this
energy (Fig. 8.8). Thus

dU=

q′q′′

4 πε 0 r

=

(4πr^2 drρ)(4πr^3 ρ/3)

4 πε 0 r

=

4 π

ε 0

ρ^2 r^4 dr

The total electrostatic energy is obtained by integrating the above expression

in the limits 0 toR,

Fig. 8.8Spherical shell of
radius R