A radioactive nucleus $n_{2}$ has 3 times the decay constant as compared to the decay constant of another radioactive nucleus $\mathrm{n}_{1}$. If initial number of both nuclei are the same, what is the ratio of number of nuclei of $\mathrm{n}_{2}$ to the number of nuclei of $\mathrm{n}_{1}$, after one half-life of $\mathrm{n}_{1}$ ?

JEE Main · 2025 · 23 Jan 2025 · Shift 1 · Modern Physics

A radioactive nucleus

n_{2}

has 3 times the decay constant as compared to the decay constant of another radioactive nucleus

n_{1}

. If initial number of both nuclei are the same, what is the ratio of number of nuclei of

n_{2}

to the number of nuclei of

n_{1}

, after one half-life of

n_{1}

Answer: (A)

The number of undecayed nuclei follows

N (t) = N_{0} e^{- λ t}

. Nucleus

n_{2}

has decay constant

λ_{2} = 3 λ_{1}

, and both start with the same number of nuclei

N_{0}

.

After time

t

equal to the half-life of

n_{1}

t = \frac{ln 2}{λ _{1}}

, so

λ_{1} t = ln 2

.

Number of

n_{1}

remaining:

N_{1} = N_{0} e^{- λ_{1} t} = N_{0} e^{- l n 2} = \frac{N _{0}}{2}

.

Number of

n_{2}

remaining:

N_{2} = N_{0} e^{- 3 λ_{1} t} = N_{0} e^{- 3 l n 2} = \frac{N _{0}}{8}

.

The ratio is

\frac{N _{2}}{N _{1}} = \frac{N _{0} /8}{N _{0} /2} = \frac{1}{4}

JEE Main · 2025 · 23 Jan 2025 · Shift 1 · Single correct option