The activity of a radioactive sample is measured as 9750 counts per minute at t = 0 & 975 counts per minute at T = 5 minutes. The decay constant is approximately.

0.922 per minutes
0.691 per minutes
0.461 per minutes
0.230 per minutes

Solution

A heavy nucleus having mass number 200 gets disintegrated into two small fragments of mass number 80 and 120. If binding energy per nucleon for parent atom is 6.5 MeV and for daughter nuclei is 7 MeV and 8 MeV respectively, then the energy released in the decay is X × 110 MeV. Find the value of X.

Solution

Energy released = (80 × 7 + 120 × 8 – 200 × 6.5)

In an α-decay the kinetic energy of α-particle is 48 MeV and Q-value of the reaction is 50 MeV.The mass number of the mother nucleus is X. Find value of X/25.(Assume that daughter nucleus is in ground state)

Solution

Half lives for α and β emission of a radioactive material are 16 years and 48 years respectively. When material decays giving α and β emission simultaneously, time in which 3/4^th material decays is

29 years
24 years
64 years
12 years

Solution

At any instant, the ratio of the amount of radioactive substances is 2 : 1. If their half lives be respectively 12 and 16 hours, then after two days, what will be the ratio of the substances ?

1 : 1
2 : 1
1 : 2
1 : 4

Solution

A neutron travelling with a velocity v and kinetic energy E has a perfectly elastic head-on collision with a nucleus of an atom of mass number A at rest. The fraction of total energy retained by the neutron is approximately

[(A– 1)/(A+ 1)]²
[(A+ 1)/(A– 1)]²
[(A– 1)/A]²
[(A+ 1)/A]²

Solution

If the total binding energies of \(_{1}^{2}H,_{2}^{4}He,_{26}^{56}Fe , _{92}^{235}U\) nuclei are 2.22, 28.3, 492 and 1786 MeV respectively, identify the most stable nucleus of the following.

\(_{26}^{56}Fe\)
\(_{1}^{2}H\)
\(_{92}^{235}U\)
\(_{2}^{4}He\)

Solution

After 150 days, the activity of a radioactive sample is 5000 dps. The activity becomes 2500 dps after another 75 days.The initial activity of the sample is

20000 dps
40000 dps
7500 dps
10000 dps

Solution

A radioactive element X converts into another stable element Y.Half life of X is 2 hrs. Initially only X is present.After time t,the ratio of atoms of X and Y is found to be 1 :4, then t in hours is

2
4
between 4 and 6
6

Solution

The radioactivity of a sample is R₁ at a time T₁ and R₂ at a time T₂. If the half-life of the specimen is T, the number of atoms that have disintegrated in the time (T₁– T₂) is proportional to

(R₁T₁ – R₂T₂)
(R₁– R₂)
(R₁– R₂)/T
(R₁– R₂) T

Solution

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