Question

Suppose a particle of ionizing radiation deposits 2.00 MeV in the gas of a Geiger tube, all of which goes to creating ion pairs. Each ion pair requires 30.0 eV of energy. (a) The applied voltage sweeps the ions out of the gas in 1.00 µs. What is the current (i

Answer 1

so current is 11.33 ×
`10^(-9)` ampere

Step-by-step explanation:

Given data

radiation deposits Q = 2.00 MeV

energy e = 30.0 eV

applied voltage v = 1.00 µs

to find out

current

solution

we know that current = Q/e i.e = 2Ne/e

here Ne is given 2.00 MeV

so current = 2 × ( 2.00 / 30.0) × 1.6 × (
`10^(-19)` / 1.6 ×
`10^(-6)` )

current = 0.22333 ×
`10^(-6)` ampere

so current is 11.33 ×
`10^(-9)` ampere