The half life period of a radioactive substance is 60 days. The time taken for 7 8 t h of its original mass to disintegrate will be:

t 1 2 = 6 0 d a y s at t = 0 m m 0 → 0 a t t → m m 0 ( 1 − 7 8 ) = m 8 m 0 → 7 8 ( m m 0 ) ⇒ 1 2 3 = 1 2 ( t 6 0 ) ⇒ t 6 0 = 3 t = 180 days

The half life period of a radioactive substance is 60 days. The time taken for $\frac{7}{8} t h$ of its original mass to disintegrate will be:

Option 1 - 120 days

Option 2 - 130 days

Option 3 - 180 days

Option 4 - 20 days

2 Views|Posted 6 months ago

Asked by Shiksha User

1 Answer

Sort by:

Answered by

Vishal Baghel | Contributor-Level 10

6 months ago

Correct Option - 3

Detailed Solution:

$t_{\frac{1}{2}} = 60 d a y s$

at t = 0 $\frac{m}{m_{0}} \to$ 0

$a t t \to \frac{m}{m_{0}} (1 - \frac{7}{8})$

= $\frac{m}{8 m_{0}} \to \frac{7}{8} (\frac{m}{m_{0}})$

$\Rightarrow \frac{1}{2^{3}} = \frac{1}{2^{(\frac{t}{60})}} \Rightarrow \frac{t}{60} = 3$

t = 180 days

Upvote

Taking an Exam? Selecting a College?

Get authentic answers from experts, students and alumni that you won't find anywhere else.

On Shiksha, get access to

66K

Colleges

1.2K

Exams

6.9L

Reviews

1.8M

Answers

Learn more about...

Physics Nuclei 2025

View Exam Details

Share Your College Life Experience

Didn't find the answer you were looking for?

Search from Shiksha's 1 lakh+ Topics

Ask Current Students, Alumni & our Experts

Quick Actions

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Have a question related to your career & education?

See what others like you are asking & answering