(D) The energy released $Q$ in a nuclear reaction is given by the mass defect multiplied by the energy equivalent of $1 \, u$ $(931.5 \, MeV/c^2)$.Mas

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(D) The energy released $Q$ in a nuclear reaction is given by the mass defect multiplied by the energy equivalent of $1 \, u$ $(931.5 \, MeV/c^2)$.Mass defect $\Delta m = (m_A - m_B - m_D)$.$\Delta m = (238.05079 - 234.04363 - 4.00260) \, u$.$\Delta m = 238.05079 - 238.04623 = 0.00456 \, u$.Energy released $Q = \Delta m \times 931.5 \, MeV/u$.$Q = 0.00456 \times 931.5 \, MeV \approx 4.24764 \, MeV$.Rounding to the nearest value,$Q \approx 4.25 \, MeV$.

Class 12 Physics Nuclei Mass-Energy, Nuclear Binding Energy, Nuclear Stability

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${ }_{92}^{238} A \rightarrow{ }_{90}^{234} B +{ }_2^4 D + Q$
In the given nuclear reaction,the approximate amount of energy released will be $.....\,MeV$.
[Given: mass of ${ }_{92}^{238} A = 238.05079 \, u$,mass of ${ }_{90}^{234} B = 234.04363 \, u$,mass of ${ }_2^4 D = 4.00260 \, u$,and $1 \, u = 931.5 \, MeV/c^2$]

JEE MAIN 2023 All JEE MAIN

A
$3.82$
B
$5.9$
C
$2.12$
D
$4.25$

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Mass-Energy, Nuclear Binding Energy, Nuclear Stability

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The binding energy per nucleon of a nucleus ${}_Z X^A$ at rest is $6 \ MeV$. It undergoes $\beta^-$ decay as shown below:
${}_Z X^A \to {}_{Z+1} Y^A + {}_{-1}^0 e + \bar{\nu}$
The total kinetic energy $(K.E.)$ of the products is $3 \ MeV$. The binding energy per nucleon of $Y$ (in $MeV$) is:

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The mass defect in a particular nuclear reaction is $0.3 \,g$. The amount of energy liberated in kilowatt-hour $(kWh)$ is: (Velocity of light $c = 3 \times 10^8 \,m/s$)

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Derive the formula for the binding energy of a nucleus by explaining the mass defect of the nucleus and write the formula for the binding energy per nucleon.

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State Einstein's special theory of relativity and provide the mass-energy equivalence formula.

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If the binding energy of the deuterium is $2.23 \, MeV$, the mass defect in $a.m.u.$ is:

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The binding energy per nucleon of a nucleus ${}_Z X^A$ at rest is $6 \ MeV$. It undergoes $\beta^-$ decay as shown below:${}_Z X^A \to {}_{Z+1} Y^A + {}_{-1}^0 e + \bar{\nu}$The total kinetic energy $(K.E.)$ of the products is $3 \ MeV$. The binding energy per nucleon of $Y$ (in $MeV$) is:

The mass defect in a particular nuclear reaction is $0.3 \,g$. The amount of energy liberated in kilowatt-hour $(kWh)$ is: (Velocity of light $c = 3 \times 10^8 \,m/s$)

Derive the formula for the binding energy of a nucleus by explaining the mass defect of the nucleus and write the formula for the binding energy per nucleon.

State Einstein's special theory of relativity and provide the mass-energy equivalence formula.

If the binding energy of the deuterium is $2.23 \, MeV$, the mass defect in $a.m.u.$ is:

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The binding energy per nucleon of a nucleus ${}_Z X^A$ at rest is $6 \ MeV$. It undergoes $\beta^-$ decay as shown below:
${}_Z X^A \to {}_{Z+1} Y^A + {}_{-1}^0 e + \bar{\nu}$
The total kinetic energy $(K.E.)$ of the products is $3 \ MeV$. The binding energy per nucleon of $Y$ (in $MeV$) is: