Answer to Question #88619 in Atomic and Nuclear Physics for Sridhar

Question

500 alpha particles of energy 5 MeV enter ionization chamber per second the energy required for ion pair creation is 25 eV. the ionization currently produced is
1)1.6×10^-11
2)9×10^-11

Answer 1

Number of pairs per second is equal

$$n=\frac {500 \times 5 \times 10^6 } {25} \times 1 s =10^8 pairs/ s (1)$$

Electric current

$$i=10^8 pairs/ s \times1.6 \times 10^{-19} C \times 1 s = 1.6 \times 10^{-11} C$$

Answer:

$1.6 \times 10^{-11} C$