Question

A Geiger counter used in several applications over the course of a typical day produces on the average 100 counts per second. The tube is in the form of a cylinder 5 cm in diameter by 20 cm long and is filled with a mixture of 90% argon and 10% ethanol to a pressure of 0.1 atmosphere. In the Geiger-Muller region, each output count results from the formation of about 1010 ion-electron pairs. How long will it take for one-third of the quenching gas to be used up, thus necessitating replacement of the tube

Answer 1

A Geiger counter is a device used to detect and measure radiation. It consists of a cylindrical tube filled with an inert gas, such as argon and ethanol. When radiation enters the tube, it ionizes the gas molecules, creating ion-electron pairs that are detected as counts. The time it takes for one-third of the quenching gas to be used up can be calculated based on the number of ion-electron pairs formed per count and the average count per second.

Step-by-step explanation:

A Geiger counter is a device used to detect and measure radiation. It consists of a cylindrical tube filled with an inert gas, such as argon and ethanol. When radiation enters the tube, it ionizes the gas molecules, creating ion-electron pairs. These pairs are attracted to the wire and cylinder in the tube, producing a current that is registered as a count. Over time, the quenching gas in the tube is used up, and one-third of it will necessitate tube replacement.

In this case, the Geiger counter produces an average of 100 counts per second. Each count corresponds to the formation of approximately 10^10 ion-electron pairs. If one-third of the quenching gas is used up, it means that one-third of the gas molecules have been ionized. Therefore, to calculate the time it takes for one-third of the quenching gas to be used up, we need to determine how many ion-electron pairs are formed by 100 counts per second and how many pairs correspond to one-third of the gas.

By dividing the total number of ion-electron pairs formed by 100 counts per second, we can calculate the time it takes for one-third of the gas to be used up.

Answer 2

As it is given that

`activity=\lambda= 100\\`

`Number~of ~counts=100/second\\`

Using formula

`ln((N)/(N_o))=-\lambda\cdot t\\\\`

or

`-(ln((N)/(N_o)))/(\lambda)= t\\\\`

where
`N=(N_o)/(3)`

`-(ln((1)/(3)))/(100)= t\\\\\-(-1.0986)/(100)=t\\\\`

`(1.0986)/(100)=t\\1.0986\cdot 10^(-2)seconds=t\\`