Question

A freshly prepared radioactive source of half-life 2h emits radiations of intensity which is 64 times the permissible safe level. The minimum time after which it would be possible to work safely with this source is-
(a) 6h
(b) 12h
(c) 24h
(d) 28h

Answer 1

Use the first-order decay law, from the half-life period equation calculate the disintegration constant and put it in the decay law equation to calculate the time i.e.
$N = {N_ \circ }{e^{ - \lambda t}}$
Where N= nuclei at any time t
${N_ \circ }$=initial number of nuclei
$\lambda$= disintegration constant

Complete step by step answer:
Step1: Using the first order decay law equation we have,
$N = {N_ \circ }{e^{ - \lambda t}}$ ………(1)
Where N= nuclei at any time t
${N_ \circ }$=initial number of nuclei
$\lambda$= disintegration constant

Step2: From half life period equation we have,
${t_{\dfrac{1}{2}}} = \dfrac{{\ln 2}}{\lambda }$
Given that ${t_{\dfrac{1}{2}}}$=2h
Therefore, $\lambda = \dfrac{{\ln 2}}{2}$ ……..(2)

Step3: Given that intensity of radiations is 64 times of permissible
Therefore, $\dfrac{N}{{{N_ \circ }}}$=$\dfrac{1}{{64}}$
Now substitute all the values in equation (1) we have,
$\dfrac{1}{{64}} = {e^{ - \dfrac{{\ln 2}}{2}t}}$
Taking log on both sides-
$\ln \left( {\dfrac{1}{{64}}} \right) = \ln \left( {{e^{ - \dfrac{{\ln 2}}{2}t}}} \right)$
Using the property of logarithm $\ln \left( {\dfrac{a}{b}} \right) = \ln \left( a \right) - \ln \left( b \right)$
$\log {}_ee = \ln e$ and
$\ln {e^a} = a$
$\ln 1 - \ln 64 = - \dfrac{{\ln 2}}{2}t$
$0 - \ln {2^5} = - \dfrac{{\ln 2}}{2}t$ $(\because \ln 1 = 0)$
$\Rightarrow - 5\ln 2 = - \dfrac{{\ln 2}}{2}t$
(Cancelling $\ln 2$ on both sides)
$\Rightarrow t = 10h$

So total minimum time after which it will be safe =10+2=12h. Hence option (B) is the correct option.

Additional information:
Half-life is the interval of time required for one-half of the atomic nuclei of a radioactive sample to decay i.e. change spontaneously into other nuclear species by emitting particles and energy or it may also be defined as the time interval required for the number of disintegrations per second of a radioactive element.
Also disintegration constant or Decay constant, the proportionality between the size of a population of radioactive atoms and the rate at which the population decreases because of radioactive decay.

Note:
Don’t make mistakes while solving the logarithmic expression: $\ln \left( {\dfrac{1}{{64}}} \right) = \ln \left( {{e^{ - \dfrac{{\ln 2}}{2}t}}} \right)$ and try to keep in mind the formulae of basic log properties, it will help in solving the equation easily. Also do remember the logarithm properties:
The property of logarithm $\ln \left( {\dfrac{a}{b}} \right) = \ln \left( a \right) - \ln \left( b \right)$
$\log {}_ee = \ln e$ and $\ln {e^a} = a$