Multiple Choice QuestionsA nucleus has 
mass represented by M (A, Z). If Mp and Mn denote the mass of proton and neutron respectively and BE the binding energy (in MeV), then:
BE = [M(A,Z)-ZMp - (A-Z)Mn]c2
BE = [ZMp + (A-Z)Mn -M(A,Z)]c2
BE = [ZMp + AMn - M (A,Z)]c2
BE = [ZMp + AMn - M (A,Z)]c2
B.
BE = [ZMp + (A-Z)Mn -M(A,Z)]c2
In the case of formation of a nucleus, the evolution of energy equal to the binding energy of the nucleus takes place due to the disappearance of a fraction of the total mass. If the quantity of mass disappearing is ΔM, then the binding energy is
BE = ΔMc2
From the above discussion, it is clear that the mass of the nucleus must be less than the sum of the masses of the constituent neutrons and protons. We can then write.
In the following circuit the output Y for all possible input A and B is expressed by the truth table:
| A | B | Y |
| 0 | 0 | 0 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
| A | B | Y |
| 0 | 0 | 1 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 0 |
| A | B | Y |
| 0 | 0 | 1 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 0 |
| A | B | Y |
| 0 | 0 | 1 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 0 |
If the nucleus 
has a nuclear radius of about 3.6 fm, then 
would have its radius approximately as:
6.0 fm
9.6 fm
12.0 fm
12.0 fm
The total energy of an electron in the ground state of a hydrogen atom is -13.6 eV. The kinetic energy of an electron in the first excited state is:
3.4 eV
6.8 eV
13.6 eV
13.6 eV
Monochromatic light of frequency 6.0 x 1014 Hz is produced by a laser. The power emitted is 2 x 10-3 W. The number of photons emitted, on the average, by the source per second is:
5 x 1015
5 x 1016
5 x 1017
5 x 1017
The frequency of a light wave in the material is 2 x 10 Hz and wavelength is 5000 A. the refractive index of material will be:
1.40
1.50
3.00
3.00
In the energy band diagram of a material shown below, the open circles and filled circles denote holes and electrons respectively. The material is a/an,
p-type semiconductor
insulator
metal
metal
A 5 W source emits monochromatic light of wavelength 5000 A. When placed 0.5 m away, it liberates photoelectrons from a photosensitive metallic surface. when the source si moved to a distance of 1.0 m, the number of photoelectrons liberated will be reduced by a factor of:
4
8
16
16
Two radioactive substance A and B have decay constants 5λ and λ respectively. At t = 0 they have the same number of nuclei. The ratio of a number of nuclei of A to those of B will be 
after a time interval:
1/ 4λ
4λ
2λ
1/2λ
Monochromatic light of frequency 6.0 x 1014 Hz is produced by a laser. The power emitted is 2 x 10-3 W The number of photons emitted, on the average, by the source per second is:
5 x 1015
5 x 106
5 x 1017
5 x 1017
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