Questions
Plutonium has atomic mass 210 and a decay constant equal to The number of -particles emitted per second by 1 mg is:
(Avogadro’s constant = )
detailed solution
Correct option is B
Number of α-particles per second = activity A (-dN/dt)=Nλ is required, whereN=6.0 x 1023210 x 1 x 10-3 And λ=5.8 x 10-8 per sec So, A= Nλ =6.0 x 1023210 x 1 x 10-3 x 5.8 x 10-8 =1.7 x 1011.Talk to our academic expert!
Similar Questions
The half life period of a radioactive element X is same as the mean life time of another radioactive element Y. Initially both of them have the same number of atoms. Then
799 666 8865
support@infinitylearn.com
6th Floor, NCC Building, Durgamma Cheruvu Road, Vittal Rao Nagar, HITEC City, Hyderabad, Telangana 500081.
free study material
JEE Mock Tests
JEE study guide
JEE Revision Notes
JEE Important Questions
JEE Sample Papers
JEE Previous Year's Papers
NEET previous year’s papers
NEET important questions
NEET sample papers
NEET revision notes
NEET study guide
NEET mock tests
