In a sample of radioactive material, what fraction of the initial number of active nuclei will remain disintegrated after half of a half-life of the sample?

Question

In a sample of radioactive material, what fraction of the initial number of active nuclei will remain disintegrated after half of a half-life of the sample? (A)  (1/4)  (B)  (1/2√2)  (C)  (1/√2)  (D)  √2-1

Tardigrade Admin · Accepted Answer

According to radioactive decay, (N/N0)=((1/2))t / T1 / 2 According to question, t=(T1 / 2/2) Then, (N/N0)=((1/2))1 / 2=(1/√2)

Q. In a sample of radioactive material, what fraction of the initial number of active nuclei will remain disintegrated after half of a half-life of the sample?

$\frac{1}{4}$

$\frac{1}{2 2}$

$\frac{1}{2}$

$2 - 1$

According to radioactive decay, $\frac{N}{N _{0}} = (\frac{1}{2})^{t / T_{1/2}}$
According to question, $t = \frac{T _{1/2}}{2}$
Then, $\frac{N}{N _{0}} = (\frac{1}{2})^{1/2} = \frac{1}{2}$

Q. In a sample of radioactive material, what fraction of the initial number of active nuclei will remain disintegrated after half of a half-life of the sample?

41​

22​1​

2​1​

2​−1

According to radioactive decay, N0​N​=(21​)t/T1/2​ According to question, t=2T1/2​​ Then, N0​N​=(21​)1/2=2​1​

$\frac{1}{4}$

$\frac{1}{2 2}$

$\frac{1}{2}$

$2 - 1$

According to radioactive decay, $\frac{N}{N _{0}} = (\frac{1}{2})^{t / T_{1/2}}$
According to question, $t = \frac{T _{1/2}}{2}$
Then, $\frac{N}{N _{0}} = (\frac{1}{2})^{1/2} = \frac{1}{2}$