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**Hint:**The phenomenon of the spontaneous disintegration of the atomic nucleus by emitting highly penetrating radiations as well as many other particles is natural radioactivity. We know that radioactive emissions are harmful to living beings. Here it is given that the intensity is $64$ times harmful than the permissible level. So we have to find an intensity that is $64$ times lesser than the given intensity.

**Formula used**

$N = {N_0}{e^{ - \lambda t}}$(Where $N$stands for the number of atoms at a given instant, ${N_0}$stands for the initial number of atoms. $\lambda $ is a constant called the disintegration constant or the decay constant, $t$stands for the time)

${T_{\dfrac{1}{2}}} = \dfrac{{\ln 2}}{\lambda }$ (Where ${T_{\dfrac{1}{2}}}$stands for the half-life period of the radioactive substance, $\lambda $is a constant called the disintegration constant or the decay constant)

**Complete step by step solution:**

According to the law of radioactive decay, the rate of disintegration of a radioactive substance at any instant is directly proportional to the number of atoms present in that substance at that instant.

So we know that the intensity is proportional to the number of atoms present in the sample at that instant.

We know that the number of atoms at any instant can be obtained by using the formula

$N = {N_0}{e^{ - \lambda t}}$

The half-life period of the source is given by ${T_{\dfrac{1}{2}}} = 2h$

The half-life period of any element can be obtained by using the formula, ${T_{\dfrac{1}{2}}} = \dfrac{{\ln 2}}{\lambda }$

Equating both we get

$2h = \dfrac{{\ln 2}}{\lambda }$

From this we get the disintegration constant as,

$\lambda = \dfrac{{\ln 2}}{2}$

It is given that the source is $64$times intense than the permissible safe level.

Therefore, we can say that the number of atoms in the permissible safe level will be, $\dfrac{{{N_0}}}{{64}}$

Now, we can rewrite the radioactive disintegration equation as,

$\dfrac{{{N_0}}}{{64}} = {N_0}{e^{ - \dfrac{{\ln 2}}{2}t}}$

From this we can find the time for the disintegration as follows,

$\dfrac{1}{{64}} = {e^{\dfrac{{ - \ln 2}}{2}t}}$

Taking the inverse, we get

$64 = {e^{\dfrac{{\ln 2}}{2}t}}$

Taking $\ln $on both sides, we get

$\ln ({2^6}) = \dfrac{{\ln 2}}{2}t$ ($64 = {2^6}$)

This can be written as,

$6\ln 2 = \dfrac{{\ln 2}}{2}t$

From this the time can be obtained as,

$t = 2 \times \dfrac{{6\ln 2}}{{\ln 2}}$

$\Rightarrow t = 2 \times 6 = 12$

**The answer is: Option (A): $12h$**

**Note:**

Alternate method:

Let us consider ${N_0}$to be the number of atoms at permissible safe level.

It is given that the intensity of the source is $64$times the permissible safe level, so we can write the number of atoms in the source as, $64{N_0}$

It will be safer to work with the given radio-active substance when $64{N_0}$becomes ${N_0}$.

The half-life of the given source is given as,

${T_{\dfrac{1}{2}}} = 2hr$

By definition, the half-life period of any radioactive substance is defined as the time taken by the reaction to disintegrate half the number of radioactive nuclei in a given sample.

This means that after $2hrs$,$64{N_0}$will become $32{N_0}$

In this manner, we have to check how many half-life periods are required for $64{N_0}$to become ${N_0}$

$64{N_0}\mathop \to \limits^{{T_{\dfrac{1}{2}}}} 32{N_0}$

$32{N_0}\mathop \to \limits^{{T_{\dfrac{1}{2}}}} 16{N_0}$

$16{N_0}\mathop \to \limits^{{T_{\dfrac{1}{2}}}} 8{N_0}$

$8{N_0}\mathop \to \limits^{{T_{\dfrac{1}{2}}}} 4{N_0}$

$4{N_0}\mathop \to \limits^{{T_{\dfrac{1}{2}}}} 2{N_0}$

$2{N_0}\mathop \to \limits^{{T_{\dfrac{1}{2}}}} {N_0}$

From this, we get that the number of atoms will become ${N_0}$after $6$half-life periods i.e.$6{T_{\dfrac{1}{2}}}$

$ \Rightarrow 6 \times 2 = 12hrs$

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