Question 24.97: A solid copper sphere whose radius is 1.0 cm has a very thin......

THINK The increase in electric potential at the surface of the copper sphere is proportional to the increase in electric charge.

EXPRESS The electric potential at the surface of a sphere of radius R is given by V=q / 4 \pi \varepsilon_0 R , where q is the charge on the sphere. Thus, q=4 \pi \varepsilon_0 R V .The number of electrons entering the copper sphere is N = q / e, but this must be equal to (λ / 2)t, where λ is the decay rate of the nickel.

ANALYZE (a) With R = 0.010 m, when V = 1000 V, the net charge on the sphere is

q=4 \pi \varepsilon_0 R V=\frac{(0.010 \,m )(1000 \,V )}{8.99 \times 10^9 \,N \cdot m ^2 / C ^2}=1.11 \times 10^{-9} \,C.

Dividing q by e yields

N=\left(1.11 \times 10^{-9} C \right) /\left(1.6 \times 10^{-19} C \right)=6.95 \times 10^9

electrons that entered the copper sphere. So the time required is

t=\frac{N}{\lambda / 2}=\frac{6.95 \times 10^9}{\left(3.7 \times 10^8 / s \right) / 2}=38 \,s .

(b) The energy deposited by each electron that enters the sphere is E_0=100\,keV =1.6 \times 10^{-14} J. Using the given heat capacity, we note that a temperature increase of ΔT = 5.0 K = 5.0 ºC required

E =CΔT = (14 J/K)(5.0 K) = 70 J

of energy. Dividing this by E_{0 } gives the number of electrons needed to enter the sphere (in order to achieve that temperature change):

N^{\prime}=\frac{E}{E_0}=\frac{70 \,J }{1.6 \times 10^{-14}\, J }=4.375 \times 10^{15}

Thus, the time needed is

t^{\prime}=\frac{N^{\prime}}{\lambda / 2}=\frac{4.375 \times 10^{15}}{\left(3.7 \times 10^8 / s \right) / 2}=2.36 \times 10^7\, s

or roughly 270 days.

LEARN As more electrons get into copper, more energy is deposited, and the copper sample gets hotter.