Question

A radioactive substance of half-life 6mins is placed near a Geiger counter which is found to register 1024 particles per minute. How many particles per minute will be registered after 42 minutes?
$(a){\text{ 4/min}} \\\ (b){\text{ 8/min}} \\\ (c){\text{ 5/min}} \\\ (d){\text{ 7/min}} \\\$

Answer 1

Hint: In this question use the concept that radioactive decay is a first order reaction. We know the formula to calculate the amount of substrate remaining after time T and with the help of it, we can easily calculate the mass remaining after a given amount of time. The equation is $N = {N_o}{e^{\left( { - \lambda t} \right)}}$ , where N is the amount of mass remaining after a given amount of time , $\lambda$is the reaction constant and is the time at which the remaining mass is to be calculated . We know the reaction constant for a first order reaction i.e. $\lambda = \dfrac{{\ln 2}}{\tau }$. Where $\tau$ is the half-life of the isotope. This will help approaching the problem.

Complete step-by-step answer:
As we know a radioactive decay is a first order reaction and the reaction constant for the for the first order reaction is given as,
$\lambda = \dfrac{{\ln 2}}{\tau }$................. (1)
Where, $\tau$ is the half-life of the substance.
Now it is given that the Geiger counter which is found to register 1024 particles per minute and the half-life of the radioactive substance is 6 min.
Then we have to find out how many particles per minute will be registered after 42 min.
So the time (t) = 42 min.
Now the particles per minute will be register in time t is given as
$\Rightarrow N = {N_o}{e^{\left( { - \lambda t} \right)}}$................. (2)
Where, ${N_o}$ = actual particles register per minute
N = amount of particles register per minute after t.
Now substitute the value from equation (1) in equation (2) we have,
$\Rightarrow N = {N_o}{e^{\left( { - \dfrac{{\ln 2}}{\tau }t} \right)}}$
Now substitute the values of half-life, time (t) and actual particle register per minute of the substance (${N_o}$) we have,
$\Rightarrow N = 1024{e^{\left( { - \dfrac{{\ln 2}}{6} \times 42} \right)}}$
Now simplify this we have,
$\Rightarrow N = 1024{e^{\left( { - 7\ln 2} \right)}}$
Now use the logarithmic property i.e. $a\ln b = \ln {b^a}$ and ${e^{\ln x}} = x$ so use this properties in the above equation we have,
$\Rightarrow N = 1024{e^{\left( {\ln {2^{ - 7}}} \right)}}$
$\Rightarrow N = 1024\left( {{2^{ - 7}}} \right)$
$\Rightarrow N = \dfrac{{1024}}{{{2^7}}}$
Now we all know that ${2^7}$ = 128 so we have,
$\Rightarrow N = \dfrac{{1024}}{{128}} = 8$ Particles per minute.
So this is the required answer.
Hence option (B) is the correct answer.

Note – Ionizing radiations are generally measured and detected by Geiger counters in practical laboratories dealing with such radioactive materials. There can be an alternate approach to solve this problem statement involving half-life, we can use the below formula directly instead of the basic approach as discussed above. The formula is $N = \dfrac{{{N_0}}}{{{2^n}}}$ and $n = \dfrac{t}{\tau }$ , where is the amount of mass remaining after a given amount of time,${N_0}$ is the initial mass of the radioisotope, $\tau$ is the half-life.