Mass Defect and Binding Energy

The above is a plot of binding energy per nucleon $E_b$, against the nuclear mass $M_i, A, B, C, D, E$ correspond to different nuclei. Consider four reactions:

(i) $A+B\to C+\epsilon$
(ii) $C\to A+B+\epsilon$.
(iii) $D+E\to F+\epsilon$ and
(iv) $F\to D+E+\epsilon$,
where $\epsilon$ is the energy released. In which reactions $\epsilon$ positive?

${{}_6\mathrm{C}}^{11}$ undergoes a decay by emitted $\beta^{+}$ then write its complete equation Given the mass value of $m\left(6^{11}\right)=11,011434u_{1}m\left({}_5{B}^{11}\right)=11.009305u$$m_{\theta }=0.000548u$ and $1\mathrm{u}=931.5\mathrm{MeV} {\mathrm{c}}^{-2}$ Calculate the Q-value of reaction.

If $M\left(A, Z\right)$, $M_p$ and $M_n$ denote the masses of the nucleus $AX$, proton and neutron respectively in units of $u\left(1u=931.5 \mathrm{MeV} {\mathrm{c}}^{-2}\right)$ and $BE$ represents its binding energy in $\mathrm{MeV}$, then

Helium nuclei combine to form an oxygen nucleus. The energy released in the reaction is if $m_{\mathrm{O}}=15.9994 \mathrm{amu}$ and $m_{\mathrm{He}}=4.0026 \mathrm{amu}$

The dynamic mass of electron having rest mass $m_0$ and moving with speed $0.8c$ is

In a fission reaction, the average binding energy per nucleon of $X$ and $Y$ is $8.5 \mathrm{MeV}$ whereas that of ${}^{236}\mathrm{U}$ is $7.6 \mathrm{MeV}$. The total energy liberated will be about ${}^{236}\mathrm{U}⟶{}^{117}\mathrm{X}+{}^{117}\mathrm{Y}+n+n$

$M_n$ and $M_p$ represent the mass of neutron and proton, respectively. An element having mass $M$ and $N$ neutron, $Z$-protons, then the correct relation will be

The binding energy per nucleon v/s mass number curve for nuclei is shown in the figure. $W, X, Y$ and $Z$ are four nuclei indicated on the curve. The process that would release energy is:

The figure shows a plot of binding energy per nucleon $(B.E./A)$ vs mass number $\left(A\right)$ for nuclei. Four nuclei, $P, Q, R$ and $S$ are indicated on the curve. The process that would release energy is

$M_p$ and $M_N$ are masses of proton and neutron, respectively, at rest. If they combine to form a deuterium nucleus. The mass of the nucleus will be

Binding energy per nucleon is of the order of -

If $M_{\mathrm{O}}$ is the mass of an oxygen isotope ${{}_8\mathrm{O}}^{17},M_p$ and $M_N$ are the masses of a proton and a neutron, respectively, the nuclear binding energy of the isotope is 

If $M_O$ is the mass of an oxygen isotope ${{}_8\mathrm{O}}^{17}$, $M_p$ and $M_n$ are the masses of a proton and a neutron, respectively, the nuclear binding energy of the isotope is

The binding energies of two nuclei ${\mathrm{P}}^n$ and ${\mathrm{Q}}^{2n}$ and $x$ and $y$ joules. If $2x>y$ then the energy released in the reaction $P^n+P^n\to Q^{2n}$, will be

Mass-energy equation $E=m{c}^2$ was given by

$1$ amu is equivalent to

In the reaction ${}_1{}^2\mathrm{H}+{}_1{}^3\mathrm{H}\to {}_2{}^4\mathrm{He}+{}_0{}^1\mathrm{n},$ if binding energies of ${}_1{}^2\mathrm{H}, {}_1{}^3\mathrm{H}$ and ${}_2{}^4\mathrm{He}$ are respectively $\mathrm{a}, \mathrm{b}$ and $\mathrm{c}$ (in $\mathrm{MeV}$), then the energy released in this reaction is:

A fission reaction is given by ${}_{92}{}^{236}\mathrm{U}\to {}_{54}{}^{140}\mathrm{Xe}+{}_{38}{}^{94}\mathrm{Sr}+\mathrm{x}+\mathrm{y}$ where $x$ and $y$ are two particles. Considering ${}_{92}{}^{236}U$ to be at rest, the kinetic energies of the products are denoted by $K_{Xe}, K_{Sr}, K_x\left(2\mathrm{MeV}\right)$ and $K_y(2 \mathrm{MeV}),$ respectively. Let the
binding energies per nucleon of ${}_{92}{}^{236}\mathrm{U},{}_{54}{}^{140}\mathrm{Xe}$ and ${}_{38}{}^{94}\mathrm{Sr}$ be $7.5 \mathrm{MeV}, 8.5 \mathrm{MeV}$ and $8.5 \mathrm{MeV},$ respectively. Considering different conservation laws, the correct option(s) is(are)

The masses of nucleus, neutrons and protons are $M\text{,} m_{\mathrm{n}}$ and $m_{\mathrm{p}}$, respectively. If the nucleus has been divided into neutrons and protons, then,

For the stability of any nucleus,