Question:
The half-life of a certain radio isotope is \(4\) minutes. The number of radioactive nuclei at a given instant is \(10^6\). Then the number of radioactive nuclei left \(2\) minutes later would be
The half-life of a certain radio isotope is \(4\) minutes. The number of radioactive nuclei at a given instant is \(10^6\). Then the number of radioactive nuclei left \(2\) minutes later would be
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If time equals half of half-life:
\[
N=\frac{N_0}{\sqrt2}
\]
This is a very common radioactive decay shortcut.
Updated On: Oct 6, 2026
- \( \dfrac{10^6}{2} \)
- \(10^3\)
- \( \dfrac{10^6}{\sqrt{2}} \)
- \( \sqrt{2}\times10^6 \)
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The Correct Option is C
Solution and Explanation
Concept:
Radioactive decay law:
\[
N=N_0\left(\frac12\right)^{t/T_{1/2}}
\]
where:
• \(N_0\) = initial nuclei
• \(T_{1/2}\) = half life
Step 1: Substitute values. Given: \[ N_0=10^6 \] \[ T_{1/2}=4\,\text{min} \] \[ t=2\,\text{min} \] Thus: \[ N=10^6\left(\frac12\right)^{2/4} \] \[ N=10^6\left(\frac12\right)^{1/2} \] \[ N=\frac{10^6}{\sqrt2} \] Hence: \[ \boxed{\frac{10^6}{\sqrt2}} \]
• \(N_0\) = initial nuclei
• \(T_{1/2}\) = half life
Step 1: Substitute values. Given: \[ N_0=10^6 \] \[ T_{1/2}=4\,\text{min} \] \[ t=2\,\text{min} \] Thus: \[ N=10^6\left(\frac12\right)^{2/4} \] \[ N=10^6\left(\frac12\right)^{1/2} \] \[ N=\frac{10^6}{\sqrt2} \] Hence: \[ \boxed{\frac{10^6}{\sqrt2}} \]
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