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

• A. \( \dfrac{10^6}{2} \)
• B. \(10^3\)
• C. \( \dfrac{10^6}{\sqrt{2}} \)
• D. \( \sqrt{2}\times10^6 \)

Hint:

If time equals half of half-life: \[ N=\frac{N_0}{\sqrt2} \] This is a very common radioactive decay shortcut.

Answer 1

The Correct Option is C

Step 1: Use the decay law.
The number of nuclei left after some time is \[ N = N_0\left(\frac12\right)^{t/T_{1/2}} \] where $N_0$ is the starting number and $T_{1/2}$ is the half-life.

Step 2: List the values.
Here $N_0 = 10^6$, half-life $T_{1/2} = 4$ min, and time $t = 2$ min.
Step 3: Find the exponent.
\[ \frac{t}{T_{1/2}} = \frac{2}{4} = \frac12 \]
Step 4: Substitute into the law.
\[ N = 10^6\left(\frac12\right)^{1/2} \]
Step 5: Simplify the half power.
Raising one half to the power one half gives $\dfrac{1}{\sqrt2}$. \[ N = \frac{10^6}{\sqrt2} \]
Step 6: State the answer.
So after $2$ minutes the count is \[ \boxed{\dfrac{10^6}{\sqrt2}} \]