Correct Answer – Option 2 : 200

Concept:

There are two common time measures of how long any given type of radionuclide lasts. One measure is the half-life T_1/2 of a radionuclide, which is the time at which both N and R have been reduced to one-half their initial values.

The other measure is the mean life τ, which is the time at which both N and R have been reduced to e^–1of their initial values. To relate T_1/2 to the disintegration constant λ, we put R = (1/2) R₀ and t = T_1/2 in and solve for T_1/2, we find T_1/2 = ln2/λ = 0.693/λ .The average life or mean life, τ can also be obtained.

Calculation:

Given,

At t = 0, count rate or initial activity is A₀ = 1600 s^-1

At t = 8 s, count rate or activity is A = 100 s^-1

So, decay scheme for given sample is

\(1600\mathop \to \limits^{{T_{1/2}}} 800\mathop \to \limits^{{T_{1/2}}} 400\mathop \to \limits^{{T_{1/2}}} 200\mathop \to \limits^{{T_{1/2}}} 100\)

So,

8 s = 4T_1/2

Where, T_1/2 = Half-life time.

\(\Rightarrow {T_{1/2}} = \frac{8}{4}\;s\) 

∴ T_1/2 = 2 s

∴ From above decay scheme, we see that activity after 6s is 200 counts per second i.e., after every two seconds, it reduced to half of its initial value.

I.e.,

After first 2 s, it reduced from 1600 to 800

After next 2 s i.e., t = 4 s, it reduced from 800 to 400.

After next 2 s i.e., t = 6 s, it reduced from 400 to 200.

Therefore, the count rate observed as counts per second at t = 6 s is close to 200 counts per second.