# NEET Physics Nuclei Questions Solved

Two radioactive substances A and B have decay constants $5\lambda$ and $\lambda$ respectively. At t=0 they have the same number of nuclei. The ratio of number of nuclei of A to those of B will be ${\left(\frac{1}{e}\right)}^{2}$ after a time interval

(a) $\frac{1}{4\lambda }$                                          (b) $4\lambda$

(c) $2\lambda$                                            (d) $\frac{1}{2\lambda }$

(d) Number of nuclei remained after time t can be wriiten as

$N={N}_{0}{e}^{-\lambda t}$

where, ${N}_{0}$ is initial number of nuclei of both the substances.

${N}_{1}={N}_{0}{e}^{-5\lambda t}$                              ...(i)

and           ${N}_{2}={N}_{0}{e}^{-\lambda t}$                               ...(ii)

Dividing Eq. (i) by Eq. (ii), we obtain

But, we have given

$\frac{{N}_{1}}{{N}_{2}}={\left(\frac{1}{e}\right)}^{2}=\frac{1}{{e}^{2}}$

Hence,        $\frac{1}{{e}^{2}}=\frac{1}{{e}^{4\lambda t}}$

Comparing the powers, we get

Difficulty Level:

• 24%
• 20%
• 18%
• 41%
Crack NEET with Online Course - Free Trial (Offer Valid Till August 23, 2019)