Answer to Question #206427 in Atomic and Nuclear Physics for Nick

Question

Alpha particle of energy 5 MeV pass through an ionization chamber at the rate of 10 per second. Assuming all the energy is used in producing ion pairs, calculate the current produced. (35 eV is required for producing an ion pair and e = 1.6×10-19 C

Answer 1

Gives

Alpha particle enter per second ionization chamber

(N)=10

Energy(per secound) of Alpha particle in ionization chamber $E=10\times5=50MeV$

Number of Alpha  particles entered in ionization chamber (pre second)=$n=\frac{50\times10^6}{35}$

ion pair given electron charge(per second)(q)=ne

$ne=\frac{50\times10^6}{35} \times1.6\times10^{-19}C$

Ionization current

$$I=\frac{ne}{t}=\frac{ne}{1}=\frac{50\times10^6}{35}\times1.6\times 10^{-19}$$

$I=2.29\times10^{-13}A$