A radioactive element has an initial count
rate of 1200cpm which reduces to 150cpm
in 15 hrs. Calculate the half-life of the
element.
A. 3 hrs
B. 5 hrs
C. 10 hrs
D. 12 hrs



Answer :

To determine the half-life of the radioactive element, we can use the relationship for radioactive decay. The formula that relates the initial count rate [tex]\( N_0 \)[/tex], the final count rate [tex]\( N(t) \)[/tex], and the half-life [tex]\( T_{\text{half}} \)[/tex] over a given time [tex]\( t \)[/tex] is given by:

[tex]\[ N(t) = N_0 \times \left(\frac{1}{2}\right)^{\frac{t}{T_{\text{half}}}} \][/tex]

Here are the steps to find the half-life:

1. Given Data:
- Initial count rate ([tex]\( N_0 \)[/tex]): 1200 counts per minute (cpm)
- Final count rate ([tex]\( N(t) \)[/tex]): 150 counts per minute (cpm)
- Time elapsed ([tex]\( t \)[/tex]): 15 hours

2. Set up the decay formula:
[tex]\[ 150 = 1200 \times \left(\frac{1}{2}\right)^{\frac{15}{T_{\text{half}}}} \][/tex]

3. Isolate the decay factor:
[tex]\[ \left(\frac{1}{2}\right)^{\frac{15}{T_{\text{half}}}} = \frac{150}{1200} \][/tex]

4. Simplify the right-hand side:
[tex]\[ \left(\frac{1}{2}\right)^{\frac{15}{T_{\text{half}}}} = \frac{1}{8} \][/tex]

5. Rewrite [tex]\(\frac{1}{8}\)[/tex] as a power of [tex]\(\frac{1}{2}\)[/tex]:
[tex]\[ \frac{1}{8} = \left(\frac{1}{2}\right)^3 \][/tex]

6. Set the exponents equal to each other:
[tex]\[ \frac{15}{T_{\text{half}}} = 3 \][/tex]

7. Solve for [tex]\( T_{\text{half}} \)[/tex]:
[tex]\[ T_{\text{half}} = \frac{15}{3} \][/tex]
[tex]\[ T_{\text{half}} = 5 \, \text{hours} \][/tex]

Therefore, the half-life of the radioactive element is 5 hours.

Hence, the correct answer is:
B. 5 hrs