Please wait...
Contact Us
Contact
Need assistance? Contact us on below numbers

For Study plan details

10:00 AM to 7:00 PM IST all days.

For Franchisee Enquiry

OR

or

Thanks, You will receive a call shortly.
Customer Support

You are very important to us

For any content/service related issues please contact on this number

93219 24448 / 99871 78554

Mon to Sat - 10 AM to 7 PM

Radioactive material 'A' has decay constant '8 lambda ' and material B has decay constant lambda initially they have same no.  Of nuclei.  After what time,  ratio of no.  Of nuclei of material B to that A will be 1/e?
 

Asked by ay91555 12th July 2017, 1:10 PM
Answered by Expert
Answer:
The following steps are to be followed for solving the query:
 
Step 1) Since the problem deals with decay constants, use the radioactive decay formula begin mathsize 12px style straight N equals straight N subscript 0 straight e to the power of negative λt end exponent end style.
Step 2) Apply the formula for both materials A and B and find the equation NA and NB.
 
Step 3) Divide NA and NB as the ratio is given. Equate it to 1/e (given). You should get begin mathsize 12px style straight N subscript straight A over straight N subscript straight B equals 1 over straight e equals straight e to the power of negative open parentheses straight lambda subscript straight A minus straight lambda subscript straight B close parentheses straight t end exponent end style.
Step 4) Substitute decay constant for A and B and find the answer for t. You should get begin mathsize 12px style straight t equals fraction numerator 1 over denominator 8 straight lambda minus straight lambda end fraction equals fraction numerator 1 over denominator 7 straight lambda end fraction end style.
 
 
 
Answered by Expert 12th July 2017, 5:10 PM
Rate this answer
  • 1
  • 2
  • 3
  • 4
  • 5
  • 6
  • 7
  • 8
  • 9
  • 10

You have rated this answer /10

Your answer has been posted successfully!

Free related questions

Chat with us on WhatsApp