If the density of gold nucleus is X, then the density of silver nucleus will be:

1.  2X

2. $\frac{X}{3}$

3.  4X

4.  X 

A nucleus with mass number 220 initially at rest emits an $\alpha$-particle. If the Q value of the reaction is 5.5 MeV, then the KE of $\alpha$-particle is:

1.  4.4 meV

2.  5.4 MeV

3.  5.6 MeV

4.  6.5 MeV

The volume (V) of a nucleus is related to its mass (M) as:

1.  $V$ $\propto$ $M$

2.  $V$ $\propto$ $\frac{1}{M}$

3.  $V$ $\propto$ $M^3$

4.  $V$ $\propto$ $\frac{1}{M^3}$

The statement which is incorrect about nuclear force between two protons is?

Binding energy per nucleon of a fixed nucleus ^A$X$ is 8 MeV. It absorbs a neutron moving with kinetic energy 4 MeV and converts into Y emitting a photon of energy 2 MeV. The binding energy per nucleon of Y (in MeV) is:

1.  $\frac{8A+2}{A+1}$

2.  $\frac{8A-2}{A+1}$

3.  $\frac{8A-1}{A+1}$

4.  $\frac{8A}{A+1}$

After two alpha decays and four beta(-ve) decays, the atomic number:

${}_8{}^{19}\mathrm{O}$ $\to$ ${}_9{}^{19}\mathrm{F}$ $+$ $\mathrm{A}$ $+$ $\mathrm{B}$

In the given decay equation, A and B indicate:

If the nuclear density of the material of atomic mass 27 is \(3\rho _{0},\)$t$hen the nuclear density of the material of atomic mass 125 is:

1.  5${\rho }_0$

2.  3${\rho }_0$

3.  $\frac{5}{3}$${\rho }_0$

4.  ${\rho }_0$

The binding energies of the nuclei A and B are E_a and E_b respectively. If three atoms of the element B fuse to give one atom of element A and an energy Q is released, then E_a, E_b and Q are related as:

1. E_a – 3E_b = Q

2. 3E_b – E_a = Q

3. E_a + 3E_b = Q

4. E_b + 3E_a = Q

Two nuclei fuse together to form a single nucleus as ${}_1{}^2\mathrm{A}+{}_1{}^2\mathrm{A}\to$ ${}_2{}^4\mathrm{B}$. If binding energy per nucleon of A and B are respectively x and y, then the energy released in the process is: