Question

In a specimen the number of nuclei of a radioactive element at t = 0 time is 2048. If its half life is 5 hr, how many nuclei would have been disintegrated in 25 hr?

Answer 1

Dear Student,

◆ Answer -

1984 nuclei

● Explaination -

Nuclei remanining after 25 hours are given by -

N = N0.(1/2)^(t/t½)

N = 2048 × (1/2)^(25/5)

N = 2048 × 1/2^5

N = 2048 / 32

N = 64 nuclei

No of nuclei disintegrated are -

∆N = N - N0

∆N = 2048 - 64

∆N = 1984 nuclei

Therefore, 1984 nuclei are disintegrated in 25 hours.

Thanks dear...

Answer 2

Given:

• nuclei number $N_{0}=2048$
• half life, T = 5 hours

To find: number of nuclei disintegrated in 25 hours = ?

Solution:

We know that, decay constant,

$\quad\quad\lambda = \frac{0.693}{T}$

$\quad\quad\quad =\frac{0.693}{5}$

$\quad\quad\quad =0.1386$

∴ the number of nuclei remaining after 25 h is given by

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

$\to N=2048\times e^{-0.1386\times 25}$

$\to N=64$

∴ the number of nuclei disintegrated in 25 h is

$\quad N_{0}-N$

= 2048 - 64

= 1984