# NEET Physics Nuclei Questions Solved

In a radioactive substance at t = 0, the number of atoms is $8×{10}^{4}$. Its half life period is 3 years. The number of atoms $1×{10}^{4}$ will remain after interval

(a) 9 years                (b) 8 years
(c) 6 years                (d) 24 years

(a) By formula $\mathrm{N}={\mathrm{N}}_{0}{\left(\frac{1}{2}\right)}^{\mathrm{t}/\mathrm{T}}$ or ${10}^{4}=8×{10}^{4}{\left(\frac{1}{2}\right)}^{\mathrm{t}/3}$
or $\left(\frac{1}{8}\right)={\left(\frac{1}{2}\right)}^{\mathrm{t}/3}$ or ${\left(\frac{1}{2}\right)}^{3}={\left(\frac{1}{2}\right)}^{\mathrm{t}/3}⇒3=\frac{\mathrm{t}}{3}$
Hence t = 9 years.

Difficulty Level:

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