Nuclei · Question 3

Assuming the experimental mass of 612C as 12 u, the mass defect of 612C atom is MeV/c2. (Mass of proton = 1.00727 u, mass of neutron = 1.00866 u, 1 u = 931.5 MeV/c2 and c is the speed of the light in vacuum).

1. 127.5
2. 89.03
3. 272.0
4. 92.0

Two nuclei of mass number 3 combine with another nucleus of mass number 4 to yield a nucleus of mass number 10. If the binding energy per nucleon for the mass numbers 3, 4 and 10 are 5.6 MeV, 7.4 MeV and 6.1 MeV, respectively, then in the process, Mc2= MeV.

1. 6.9
2. 7.9
3. 2.2
4. 4.3

The binding energy per nucleon of 20983Bi is MeV. [Take m( 20983Bi) = 208.980388 u, mp = 1.007825 u, mn = 1.008665 u, 1 u = 931 MeV/c2]

1. 7.48
2. 7.84
3. 8.79
4. 6.94

An atom 83X is bombarded by shower of fundamental particles and in 10 s this atom absorbed 10 electrons, 10 protons and 9 neutrons. The percentage growth in the surface area of the nucleus is recorded by :

1. 150\%
2. 900\%
3. 125\%
4. 225\%

A nucleus has mass number and radius R . Another nucleus has mass number and radius R . If = 8 then R /R is :

1. 1
2. 2
3. 8
4. 0.5

Given below are two statements : In the light of the above statements, choose the correct answer from the options given below

1. Statement I is false but Statement II is true
2. Both Statement I and Statement II are true
3. Both Statement I and Statement II are false
4. Statement I is true but Statement II is false

The binding energies for the following nuclear reactions are expressed in MeV. 2He3 + 0n1 2He4 + 20 MeV 2He4 + 0n1 2He5 - 0.9 MeV If X3, X4, X5 denote the stability of 2He3, 2He4 and 2He5, respectively, then the correct order is :

1. X4 > X5 > X3
2. X4 < X5 < X3
3. X4 > X5 < X3
4. X4 = X5 = X3

Which of the following pair of nuclei are isobars of the element?

1. 31H and 32He
2. 19880Hg and 19779Au
3. 21H and 31H
4. 23692U and 23892U

The average energy released per fission for the nucleus of 23592U is 190 MeV. When all the atoms of 47 g pure 23592U undergo fission process, the energy released is 1023 MeV. The value of is . (Avogadro Number = 6 1023 per mole)

The minimum frequency of photon required to break a particle of mass 15.348 amu into 4 particles is kHz. [mass of He nucleus = 4.002 amu, 1 amu = 1.66 10-27 kg, h = 6.6 10-34 J s and c = 3 108 m/s]

1. 14.94 1020
2. 9 1019
3. 9 1020
4. 14.94 1019