Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

A sample of radioactive material a that has an activity of NMC has twice the number of nuclear as other sample of a different radioactive material be which has an activity of 20 MCI the correct choices for half life of a and b would be then respectively

A sample of radioactive material a that has an activity of NMC has twice the number of nuclear as other sample of a different radioactive material be which has an activity of 20 MCI the correct choices for half life of a and b would be then respectively

Grade:12

1 Answers

Arun
25763 Points
one year ago
Law of radioactivity -
 
-\frac{dN}{dt}= \lambda N
 
- wherein
 
Ratio of disintegration is propotional to number of nuclei         
 
\lambda= disintegration constant
 
 
 
Activity A = \lambda N
 
For A , A = 2 No \lambda_{A} = 10
 
For B , A = No \lambda_{B} = 20
 
 
 
\frac{10}{20} = \frac{2 \lambda_{A}}{\lambda_{B}}
 
\lambda_{B} = 4 \lambda_{A}
 
t_{\frac{1}{2} }= \frac{ln 2 }{\lambda}
 
\frac{t_{\frac{1}{2}A }}{t_{\frac{1}{2}B }}= \frac{\lambda_{B}}{\lambda_{A}} = \frac{\lambda_{B}}{\frac{\lambda_{B}}{4}} = 4
 
t_{\frac{1}{2}A } = 4 t_{\frac{1}{2}B }
 
So option 20 days and 5 days satisfies this condition .
 
 

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free