A radio-active material is reduced to 1 / 8 of its original amount in 3 days. If 8 × 10^ -3 kg of the material is left after 5 days. The initial amount of the material is

Correct answer: D. 256 gm N=N_0((1)/(2))^ n N= N_0 8 N_0 8 =N_0((1)/(2))^ n ((1)/(2))^3=((1)/(2))^ n n=3 3 half lives =3 days 1 half life =1 day 5 days =5 half life N=N_0((1)/(2))^ n 8 × 10^ -3 =N_0((1)/(2))^5 N_0=2^5 × 8 × 10^ -3 =256 gm

BITSAT 2023Paper 1

Exam: BITSAT
Year: 2023
Paper: Paper 1
Subject: Physics
Chapter: Nuclear Physics

BITSAT 2023 · Paper 1 · Physics · Nuclear Physics

A radio-active material is reduced to 1 / 8 of its original amount...

A radio-active material is reduced to

1 / 8

of its original amount in 3 days. If

8 \times 10^{-3} \mathrm{ kg}

of the material is left after 5 days. The initial amount of the material is

Options

A $700 \mathrm{gm}$
B $900 \mathrm{gm}$
C $475 \mathrm{gm}$
D $256 \mathrm{gm}$

Correct answer

D. $256 \mathrm{gm}$

Step-by-step solution

\mathrm{N}=\mathrm{N}_0\left(\frac{1}{2}\right)^{\mathrm{n}}

\mathrm{N}=\frac{\mathrm{N}_0}{8}

\frac{\mathrm{N}_0}{8}=\mathrm{N}_0\left(\frac{1}{2}\right)^{\mathrm{n}} \Rightarrow\left(\frac{1}{2}\right)^3=\left(\frac{1}{2}\right)^{\mathrm{n}}

\mathrm{n}=3

3 half lives

=3

days
1 half life

=1

day
5 days

=5

half life

\mathrm{N}=\mathrm{N}_0\left(\frac{1}{2}\right)^{\mathrm{n}} \Rightarrow 8 \times 10^{-3}=\mathrm{N}_0\left(\frac{1}{2}\right)^5

\Rightarrow \quad N_0=2^5 \times 8 \times 10^{-3}=256 \mathrm{gm}

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