The disintegration rate of a certain radioactive sample at any instant is 4250 disintegrations per minutes later, the rate becomes 2250 disintegrations per minute. The approximate decay constant is:
(Take log10 1.88 = 0.274)
The disintegration rate of a certain radioactive sample at any instant is 4250 disintegrations per minutes later, the rate becomes 2250 disintegrations per minute. The approximate decay constant is:
(Take log10 1.88 = 0.274)
Option 1 -
0.02 min-1
Option 2 -
2.7 min-1
Option 3 -
0.063 min-1
Option 4 -
6.3 min-1
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1 Answer
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Correct Option - 3
Detailed Solution:
Similar Questions for you
Q = [4 *4.0026 – 15.9994] *931.5 MeV
Q = 10.2 MeV
-(1)
for B,
for B,
-(2)
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.
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