Composition and Size of Nucleus

Two nuclei have mass numbers in the ratio $27:125$. The ratio of their nuclear radii would be:

Two nuclei have mass numbers in the ratio $8:125$. The ratio of their nuclear radii is:

Assuming the nuclei to be spherical in shape how does the surface area of a nucleus of mass number  $A_1$  compare with that of a nucleus of mass number $A_2$ ?

The plot of the binding energy per nucleon versus the mass number A for a large number of nuclei,  $2\le A\le 240$  is shown in Fig. In which range the binding energy per nucleon is constant?

The plot of the binding energy per nucleon versus the mass number A for a large number of nuclei,  $2\le A\le 240$  is shown in Fig. In which range the binding energy per nucleon is constant?

Calculate the energy released in MeV in the following nuclear reaction :<

${}_{92}^{238}\text{U}\to {}_{90}^{234}\text{Th}+{}_2^4\text{He}+\text{Q}$

[Mass of  ${}_{92}^{238}\text{U}$ = 238.05079 u]

Mass of  ${}_{90}^{234}\text{Th}$ = 234.043630 u

Mass of  ${}_2^4\text{He}$  = 4.002600 u

1 u = 931.5 MeV

$\text{Mass of neutron}=1.008665 \mathrm{u}$

The binding energies per nucleon for deuteron (${}_1\mathrm{H}^2$) and helium (${}_2\mathrm{He}^4$) are $1.1 \mathrm{MeV}$ and $7.0 \mathrm{MeV}$ respectively. The energy released when two deuterons fuse to form a helium nucleus (${}_2\mathrm{He}^4$) is

In process of nuclear fission of $1 \mathrm{g}$, the mass lost is $0.90 \mathrm{mg}$. The efficiency of power house's fission reactor is $20\%$. To obtain$400$mega watt power from the power house how much uranium required per hour

Convert $1 \mathrm{amu}$ mass into megaelectron volt.

A nuclear fusion reaction is given by$4\left({}_2{}^{4}He\right)\to {}_8{}^{16}O$. If atomic mass of He is$4.0026 \mathrm{amu}$ and that of $O$ is $15.9994$ amu then the Q-value of the reaction will be approximately

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R

Assertion A : The binding energy per nucleon is practically independent of the atomic number for nuclei of mass number in the range $30$ to $170$.

Reason R : Nuclear force is short ranged.

In the light of the above statements, choose the correct answer from the options given below

A nucleus with mass number $242$ and binding energy per nucleon as $7.6 \mathrm{MeV}$ breaks into two fragment each with mass number $121$. If each fragment nucleus has binding energy per nucleon as $8.1 \mathrm{MeV},$ the total gain in binding energy is $\_\_\_\_\_\_\_ \mathrm{MeV}$.

For a nucleus ${}_Z{}^{A}X$ having mass number A and atomic number $\mathrm{Z}$

A. The surface energy per nucleon$\left(b_s\right)=–a_1{A}^{\frac{2}{3}}.$

B. The Coulomb contribution to the binding energy ${\mathrm{b}}_{\mathrm{c}}=-{\mathrm{a}}_2\frac{Z(Z-1)}{A^{{\displaystyle \frac{4}{3}}}}$.

C. The volume energy $b_v = a_{3}A$

D. Decrease in the binding energy is proportional to surface area.

E. While estimating the surface energy, it is assumed that each nucleon interacts with 12 nucleons. ($a_1, a_2$ and $a_3$ are constants)

Choose the most appropriate answer from the options given below:

For the given radioactive decay ${}_{94}{}^{298}X\to {}_{92}{}^{294}Y+{}_2{}^4\alpha +Q-\mathrm{value}$, binding energy per nucleon of $X,Y$ and $\alpha$ are $a, b$ and $c$. The $Q-\mathrm{value}$ is equal to

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R

Assertion A : The nuclear density of nuclides ${}_5{}^{10}B, {}_3{}^{6}Li, {}_{26}{}^{56}Fe, {}_{10}{}^{20}Ne \text{and} {}_{83}{}^{209}Bi$ can be arranged as ${\rho }_{Bi}^N>{\rho }_{Fe}^N>{\rho }_{Ne}^N>{\rho }_B^N>{\rho }_{Li}^N$

Reason R : The radius $R$ of nucleus is related to its mass number $A$ as $R=R_0 A^{\frac{1}{3}}$, where $R_0$ is a constant.

In the light of the above statement, choose the correct answer from the options given below :

Find the energy equivalent of one atomic mass unit, first in Joule and then in MeV. Using this, express the mass defect of ${}_8{}^{16}\mathrm{O}$ in $\mathrm{MeV}/c^2$.

Binding energy of a Nitrogen nucleus $\left[{}_7{}^{14} \mathrm{N}\right]$, given $m\left[{}_7{}^{14} \mathrm{N}\right]=14.00307 \mathrm{u}$