Search Colleges, Courses, Exams, Questions and Articles

Ask a query on 8879100846 Ask Login | Sign Up

 

 

› Ask & Answer › Question

A radioactive simple is undergoing a decay. At any time t1, its activity is A and another time t2, the activity is A 5 .  What is the average life time for the sample?

Option 1 -

t 2 − t 1 I n 5

Option 2 -

I n ( t 2 + t 1 ) 2

Option 3 -

t 1 − t 2 I n 5

Option 4 -

I n 5 t 2 − t 1

0 2 Views | Posted a month ago
Asked by Shiksha User

  • 1 Answer

  • A

    Answered by

    alok kumar singh | Contributor-Level 10

    a month ago
    Correct Option - 1


    Detailed Solution:

    A = A 0 e − λ t 1     [Radio active decay law]

    A 5 = A 0 e − λ ( t 2 − t 1 )

    ⇒ A v e r a g e     l i f e = 1 λ = ( t 2 − t 1 ) l n 5  

              

Similar Questions for you

The distance of closest approach of an alpha particle fired towards gold nucleus with kinetic energy k is r0. The distance of closest approach when the alpha particle is fired towards same nucleus with kinetic energy 4k will be
A
alok kumar singh

Kindly go through the solution

 

Find the mass number of an atom whose radius is half of that of a given atom of mass number 192.
A
alok kumar singh

r = R 0 ( 1 9 2 ) 1 3

r 2 = R 0 ( m ) 1 3

m = 1 9 2 8 = 2 4

A nuclear fusion reaction is given by   If atomic mass of He is 4.0026 amu and that of O is 15.9994 amu then the Q-value of the reaction will be approximately
A
alok kumar singh

Q = [4 *4.0026 – 15.9994] *931.5 MeV

Q = 10.2 MeV

A sample contains 10-2kg each of two substances A and B with half lives 4s and 8s respectively. The ration of their atomic weights is 1 : 2. The ratio of the amounts of A and B after16s is x 1 0 0 . The value of x is ______.
A
alok kumar singh

f o r     A , t 1 2 = 4     s e c  

= m A 0 e − 0 . 6 9 3 4 × 1 6

m A = m A 0 e − 4 × 0 . 6 9 3    -(1)

for B,

for B,  t 1 2 = 8 s e c  

m B = m B 0 e − 2 × 0 . 6 9 3               -(2)

m A m B = m A O m B O e − 4 × 0 . 6 9 3 e − 2 × 0 . 6 9 3

m A m B = 2 5 1 0 0 = x 1 0 0 x = 2 5

A nucleus with mass number 240 breaks into two fragments each of mass number 120, the binding energy per nucleon of unfragmented nuclei is 7.6MeV while that of fragments is 8.5MeV. The total gain in the Binding Energy in the process is:
A
alok kumar singh

The reaction is X²? → Y¹²? + Z¹²?
Binding energies per nucleon are: X=7.6 MeV, Y=8.5 MeV, Z=8.5 MeV.
Gain in binding energy (Q) = (Binding energy of products) - (Binding energy of reactants)
Q = (120 × 8.5 + 120 × 8.5) - (240 × 7.6) MeV
Q = (2 × 120 × 8.5) - (240 × 7.6) MeV = 2040 - 1824 = 216 MeV.

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

Sign Up on Shiksha

On Shiksha, get access to

  • 65k Colleges
  • 1.2k Exams
  • 686k Reviews
  • 1800k Answers

Learn more about...

Share Your College Life Experience

Write a Review

Didn't find the answer you were looking for?

Search from Shiksha's 1 lakh+ Topics

or

Ask Current Students, Alumni & our Experts

Ask Now
Community GuidelinesUser Points System
QuestionsDiscussionsTags
×

This website uses Cookies and related technologies for the site to function correctly and securely, improve & personalise your browsing experience, analyse traffic, and support our marketing efforts and serve the Core Purpose. By continuing to browse the site, you agree to Privacy Policy and Cookie Policy.

Need guidance on career and education? Ask our experts

Characters 0/140

The Answer must contain atleast 20 characters.

Add more details

Characters 0/300

The Answer must contain atleast 20 characters.

Keep it short & simple. Type complete word. Avoid abusive language. Next

Your Question

Edit

Add relevant tags to get quick responses. Cancel Post