Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • A source contains two phosphorous radio nuclides 32 15P (T1/2 = 14.3d) and 33 15P(T1/2 = 25.3d). Initially, 10% of the decays come from 33 15P. How long one must wait until 90% do so?

Q.13.25 A source contains two phosphorous radio nuclides _{15}^{32}\textrm{P}(T_{1/2}=14.3d) and _{15}^{33}\textrm{P}(T_{1/2}=25.3d). Initially, 10% of the decays come from _{15}^{33}\textrm{P}. How long one must wait until 90% do so?

Answers (1)

best_answer

Let initially there be N1 atoms of _{15}^{32}\textrm{P} and N2 atoms of _{15}^{33}\textrm{P} and let their  decay constants be \lambda _{1} and \lambda _{2} respectively

Since initially the activity of  _{15}^{33}\textrm{P} is 1/9 times that of _{15}^{32}\textrm{P} we have

N_{1 } \lambda_{1}=\frac{N_{2}\lambda _{2}}{9}      (i)

Let after time t the activity of  _{15}^{33}\textrm{P} be 9 times that of _{15}^{32}\textrm{P}

N_{1 } \lambda_{1}e^{-\lambda _{1}t}=9N_{2}\lambda _{2}e^{-\lambda _{2}t}    (ii)

Dividing equation (ii) by (i) and taking the natural log of both sides we get

\\-\lambda _{1}t=ln81-\lambda _{2}t \\t=\frac{ln81}{\lambda _{2}-\lambda _{1}}

where \lambda _{2}=0.048/ day and\lambda _{1}=0.027/ day

t comes out to be 208.5 days

Posted by

Sayak

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE board exams 2025 from February 15; Class 10, 12 subject-wise marks distribution
anam-khan

CBSE Date Sheet 2025: Class 10, 12 board exam dates soon on cbse.gov.in; previous trends, sample papers
anam-khan

CBSE Board Exams 2025: 75% attendance must for students to be eligible, reiterates board

Latest Question