Answer

Verified

51.6k+ views

**Hint:**In this question the rate of reaction for P and Q is the same that means the disintegration constant for nuclei P and Q are equal. For radioactive disintegration the change of number of nuclei is differentiated with respect to time.

**Complete step by step answer:**

Let us assume that disintegration constant is $\lambda $. Since the rate of disintegration is same for P and Q so we can write,

$ \Rightarrow {\lambda _1}{N_1} = {\lambda _2}{N_2}$

We can write the above expression as,

$ \Rightarrow {\lambda _1}{N_0}{e^{ - {\lambda _1}t}} = {\lambda _2}{N_0}{e^{ - {\lambda _2}t}}$

Now, we divide both sides of the above equation by ${\lambda _2}$.

$ \Rightarrow \dfrac{{{\lambda _1}}}{{{\lambda _2}}} = {e^{{\lambda _1}t}} - {e^{ - {\lambda _2}t}}$

After simplification of the above equation we get,

$ \Rightarrow t = \dfrac{{\ln \dfrac{{{\lambda _1}}}{{{\lambda _2}}}}}{{{\lambda _1} - {\lambda _2}}}$

And now in terms of the half life ${T_1}$ and ${T_2}$, we can write the equation as,

$ \Rightarrow t = \dfrac{{\ln \dfrac{{{T_2}}}{{{T_1}}}}}{{\ln 2\left( {\dfrac{1}{{{T_1}}} - \dfrac{1}{{{T_2}}}} \right)}}$

Now, we substitute the values in the above equation that is ${T_1} = 1\;{\text{month}}$ and ${T_2} = 2\;{\text{months}}$.

$ \Rightarrow t = \dfrac{{\ln 2}}{{\ln 2\left( {1 - \dfrac{1}{2}} \right)}}$

After simplification of the above equation we get,

$t = 2\;{\text{month}}$

In two months, P disintegrates 2 half-lives that is $75\% $.

Now we convert $75\% $ Into fraction as

$ \Rightarrow 75\% x = \dfrac{{3x}}{4}$

In two months, Q disintegrates to 1 half-life, so it is $50\% $.

Now we convert $50\% $ into fraction as,

$ \Rightarrow 50\% x = \dfrac{x}{2}$

As we know that the amount of R formed can be calculated by adding the disintegration of P and the disintegration of Q.

Add disintegration of P and disintegration of Q.

$ \Rightarrow {\text{Amount of R}} = \dfrac{{3x}}{4} + \dfrac{x}{4}$

After simplification we get,

$ \Rightarrow {\text{Amount of R}} = \dfrac{{5x}}{4}$

**Therefore, the number of nuclei of R is $\dfrac{{5x}}{4}$.**

**Note:**In this question one thing should be noted that the rate of disintegration of P and Q is the same. The amount of R formed by disintegration of P and disintegration of Q is added in this case. Don’t forget to convert the percentage disintegration into fractions for getting the answer in terms of $x$.

Recently Updated Pages

A small illuminated bulb is at the bottom of a tank class 12 physics JEE_Main

A metal wire is subjected to a constant potential difference class 12 physics JEE_main

Deflection in the galvanometer A Towards right B Left class 12 physics JEE_Main

Calculate the resistivity of the material of a wire class 12 physics JEE_Main

When an external voltage V is applied across a semiconductor class 12 physics JEE_Main

When a Voltmeter connected across the terminals of class 12 physics JEE_Main

Other Pages

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main

In Bohrs model of the hydrogen atom the radius of the class 12 physics JEE_Main

When white light passes through a hollow prism then class 11 physics JEE_Main

The resultant of vec A and vec B is perpendicular to class 11 physics JEE_Main

Explain the construction and working of a GeigerMuller class 12 physics JEE_Main