MHT CET · Physics · Nuclear Physics
A radioactive substance has half-life of 60 minute. During 3 hour, the amount of substance decayed would be
- A \(8.5 \%\)
- B \(12.5 \%\)
- C \(25 \%\)
- D \(87.5 \%\)
Answer & Solution
Correct Answer
(D) \(87.5 \%\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & N=N_0\left(\frac{1}{2}\right)^{\frac{t}{T_1 / 2}} \\ \therefore & \frac{N}{N_0} \\ \therefore & \left(\frac{1}{2}\right)^{\frac{t}{T_1 / 2}}\end{aligned}\)
\(\begin{array}{ll}
\therefore & \frac{\mathrm{N}}{\mathrm{~N}_0}=\left(\frac{1}{2}\right)^{\frac{3 \times 60}{60}}=\left(\frac{1}{2}\right)^3 \\
& \therefore \quad \\
\therefore & \frac{\mathrm{~N}}{\mathrm{~N}_0}=\frac{1}{8}
\end{array}\)
The amount of substance decayed \(=1-\frac{N}{N_0}\)
\(1-\frac{\mathrm{N}}{\mathrm{~N}_0}=1-\frac{1}{8}=\frac{7}{8}=87.5 \%\)
\(\begin{array}{ll}
\therefore & \frac{\mathrm{N}}{\mathrm{~N}_0}=\left(\frac{1}{2}\right)^{\frac{3 \times 60}{60}}=\left(\frac{1}{2}\right)^3 \\
& \therefore \quad \\
\therefore & \frac{\mathrm{~N}}{\mathrm{~N}_0}=\frac{1}{8}
\end{array}\)
The amount of substance decayed \(=1-\frac{N}{N_0}\)
\(1-\frac{\mathrm{N}}{\mathrm{~N}_0}=1-\frac{1}{8}=\frac{7}{8}=87.5 \%\)
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