Plutonium has atomic mass 210 and a decay constant equal to 5.8 x 10-8 sec-1. The number of α-particles emitted per second by 1 mg is:(Avogadro’s constant = 6.0 x 1023)
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a
1.7x 109
b
1.7x 1011
c
2.9x 1011
d
3.4 x 109
answer is B.
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Detailed Solution
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.