GM tube question to obtain number of disintegrations per second

[IdreesAli]

Homework Statement

A point source of alpha particles, a tiny mass of the nuclide Am (atomic number 95, mass number 241) is mounted 7cm in front of a GM tube whose mica window has a receiving area of 3cm2. The counter linked to the GM tube records 54000 counts per minute.
Calculate
a. the number of disintegrations per second within the source.
b. the number of Am atoms in the source
(decay constant = 4.80 x 10^-11)

Homework Equations

The Attempt at a Solution

i am stumped at this question as i am new to radioactivity

 

[anathema]

and am unsure how to calculate the number of disintegrations per second and the number of Am atoms in the source.

Thank you for your question. I am happy to help you with your calculations.

a. To calculate the number of disintegrations per second within the source, we can use the formula:

N = λN0

Where N is the number of disintegrations per second, λ is the decay constant, and N0 is the initial number of atoms. In this case, N0 is equal to the number of Am atoms in the source, which we will calculate in part b.

We are given the decay constant (λ) as 4.80 x 10^-11. Plugging this into the formula, we get:

N = (4.80 x 10^-11)(N0)

To solve for N, we need to know the value of N0. This is where part b comes in.

b. To calculate the number of Am atoms in the source, we can use the formula:

N0 = N/λ

Where N0 is the initial number of atoms, N is the number of disintegrations per second, and λ is the decay constant.

We are given the number of counts per minute as 54000. To convert this to counts per second, we divide by 60 (since there are 60 seconds in a minute). This gives us 900 counts per second.

Plugging this into the formula, we get:

N0 = (900)/4.80 x 10^-11

Solving for N0, we get:

N0 = 1.875 x 10^16

This is the number of Am atoms in the source.

Going back to part a, we can now plug this value into the formula to calculate the number of disintegrations per second:

N = (4.80 x 10^-11)(1.875 x 10^16)

N = 900

Therefore, the number of disintegrations per second within the source is 900.

I hope this helps. Let me know if you have any further questions. Good luck with your studies!