Question

A particle of ionizing radiation creates 4000 ion pairs in the gas inside a Geiger tube as it passes through. What minimum energy was deposited, if 30.0 eV is required to create each ion pair?

a. Min Energy = 0.12 , MeV
b . Min Energy = 0.08 , MeV
c .Min Energy = 0.10 , MeV
d .Min Energy = 0.06 , MeV

Answer 1

The minimum energy deposited by a particle of ionizing radiation that creates 4000 ion pairs in a Geiger tube is calculated by multiplying the number of ion pairs by the energy needed for each pair. This gives us 120,000 eV, which converts to 0.12 MeV.

Step-by-step explanation:

To calculate the minimum energy deposited when a particle of ionizing radiation creates 4000 ion pairs in the gas inside a Geiger tube, we need to multiply the number of ion pairs by the energy required to create each pair. Since it requires 30.0 eV to create one ion pair, for 4000 ion pairs, the calculation would be:

Minimum Energy = Number of ion pairs × Energy per ion pair

Minimum Energy = 4000 ion pairs × 30.0 eV/ion pair

Minimum Energy = 120,000 eV

Since 1 MeV = 1,000,000 eV, we need to convert the energy from electron volts (eV) to megaelectron volts (MeV):

Minimum Energy = 120,000 eV / 1,000,000 eV/MeV = 0.12 MeV

The correct answer to the question is a. Min Energy = 0.12 MeV.