Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

Two radioactive nuclei P and Q, in a given sample decay into a stable nucleus R. At time t = 0, the number of P species are 4N_0 and that of Q is N_0. Half-life of P (for conversion to R) is 1 min whereas that of Q is 2 min. Initially there are no nuclei of R present in the sample. When number of nuclei of P and Q are equal, the number of nuclei of R present in the sample would be:

3N_0

9N_0/2

5N_0/2

2N_0

Step-by-Step Solution

Decay Equation: The number of nuclei remaining $N$ after time $t$ is given by $N = N_{initial} \left(\frac{1}{2}\right)^{t/T_{1/2}}$ .
Find Time ( $t$ ) for Equal Nuclei:

For P: $N_P = 4N_0 \left(\frac{1}{2}\right)^{t/1}$
For Q: $N_Q = N_0 \left(\frac{1}{2}\right)^{t/2}$
Equating $N_P = N_Q$ : $4N_0 \left(\frac{1}{2}\right)^{t} = N_0 \left(\frac{1}{2}\right)^{t/2}$ $4 = \frac{(1/2)^{t/2}}{(1/2)^t} = \left(\frac{1}{2}\right)^{-t/2} = 2^{t/2}$ $2^2 = 2^{t/2} \implies \frac{t}{2} = 2 \implies t = 4 \text{ min}$

Calculate Nuclei of R Formed:

Nuclei of R formed = (Decayed P) + (Decayed Q).
Remaining P at 4 min: $N_P(4) = 4N_0 (1/2)^4 = 4N_0/16 = N_0/4$ .
Decayed P = $4N_0 - N_0/4 = 15N_0/4$ .
Remaining Q at 4 min: $N_Q(4) = N_0 (1/2)^2 = N_0/4$ .
Decayed Q = $N_0 - N_0/4 = 3N_0/4$ .
Total R = $\frac{15N_0}{4} + \frac{3N_0}{4} = \frac{18N_0}{4} = \frac{9N_0}{2}$ .

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