147477
Binding energy per nucleon versus 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
1 $\mathrm{Y} \rightarrow 2 \mathrm{Z}$
2 $\mathrm{W} \rightarrow \mathrm{X}+\mathrm{Z}$
3 $\mathrm{W} \rightarrow 2 \mathrm{Y}$
4 $\mathrm{X} \rightarrow \mathrm{Y}+\mathrm{Z}$
Explanation:
C Energy is released is a process when total binding energy of products is more than the reactants. In process- $\mathrm{W} \rightarrow 2 \mathrm{Y}$ Binding energy of reactants $W,\left|E_{W}\right|=120 \times 7.5=900$ $\mathrm{MeV}$ Binding energy of products $2 \mathrm{Y},\left|\mathrm{E}_{2 \mathrm{Y}}\right|=2(60 \times 8.5)=$ $1020 \mathrm{MeV}$ Here, $1020 \mathrm{MeV}>900 \mathrm{MeV}$ Hence, option (c) is correct.
AIIMS-2016
NUCLEAR PHYSICS
147478
For mass defect of $0.3 \%$, the binding energy of $1 \mathrm{~kg}$ material is
1 $2.7 \times 10^{14} \mathrm{erg}$
2 $2.7 \times 10^{14} \mathrm{~J}$
3 $2.7 \times 10^{-14} \mathrm{erg}$
4 $2.7 \times 10^{-14} \mathrm{~J}$
Explanation:
B Given, Mass defect $(\Delta \mathrm{m})=0.3 \%$ of $1 \mathrm{~kg}=3 \times 10^{-3} \mathrm{~kg}$ Speed $(\mathrm{c})=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ The binding energy $(E)=(\Delta \mathrm{m}) \mathrm{c}^{2}$ $\mathrm{E}=3 \times 10^{-3}\left(3 \times 10^{8}\right)^{2}$ $\mathrm{E}=27 \times 10^{13} \mathrm{~J}$ $\mathrm{E}=2.7 \times 10^{14} \mathrm{~J}$
COMEDK 2016
NUCLEAR PHYSICS
147480
When an electron-positron pair annihilates, the energy released is about
1 $0.8 \times 10^{-13} \mathrm{~J}$
2 $1.6 \times 10^{-13} \mathrm{~J}$
3 $3.2 \times 10^{-13} \mathrm{~J}$
4 $4.8 \times 10^{-13} \mathrm{~J}$
Explanation:
B Rest mass of electron and positron $(\mathrm{m})=9.1$ $\times 10^{-31} \mathrm{~kg}$ During annihilation rest mass will convert into energy. Then, $\mathrm{E}=\Delta \mathrm{mc}^{2}$ $\mathrm{E}=2 \mathrm{mc}^{2}$ $\mathrm{E}=2 \times 9.1 \times 10^{-31} \times\left(3 \times 10^8\right)^2$ $\mathrm{E}=2 \times 9.1 \times 10^{-31} \times 9 \times 10^{16} \mathrm{~J}$ So, \(\quad E=1.638 \times 10^{-13} \mathrm{~J}\)
AIIMS-2004
NUCLEAR PHYSICS
147482
The binding energy per nucleon of ${ }^{16} \mathrm{O}$ is 7.97 $\mathrm{MeV}$ and that of ${ }^{17} \mathrm{O}$ is $7.75 \mathrm{MeV}$. The energy in $\mathrm{MeV}$ required to remove a neutron from ${ }^{17} \mathrm{O}$ is
1 3.52
2 3.64
3 4.23
4 7.86
5 1.68
Explanation:
C The binding energy per nucleon for $\mathrm{O}^{16}=$ $7.97 \mathrm{MeV}$ The binding energy per nucleon for $\mathrm{O}^{17}=7.75 \mathrm{MeV}$ Energy required $=$ Binding energy of $\mathrm{O}^{17}-$ Binding energy of $\mathrm{O}^{16}$ $=17 \times 7.75-16 \times 7.97$ $=131.75-127.52$ Energy required $=4.23 \mathrm{MeV}$
147477
Binding energy per nucleon versus 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
1 $\mathrm{Y} \rightarrow 2 \mathrm{Z}$
2 $\mathrm{W} \rightarrow \mathrm{X}+\mathrm{Z}$
3 $\mathrm{W} \rightarrow 2 \mathrm{Y}$
4 $\mathrm{X} \rightarrow \mathrm{Y}+\mathrm{Z}$
Explanation:
C Energy is released is a process when total binding energy of products is more than the reactants. In process- $\mathrm{W} \rightarrow 2 \mathrm{Y}$ Binding energy of reactants $W,\left|E_{W}\right|=120 \times 7.5=900$ $\mathrm{MeV}$ Binding energy of products $2 \mathrm{Y},\left|\mathrm{E}_{2 \mathrm{Y}}\right|=2(60 \times 8.5)=$ $1020 \mathrm{MeV}$ Here, $1020 \mathrm{MeV}>900 \mathrm{MeV}$ Hence, option (c) is correct.
AIIMS-2016
NUCLEAR PHYSICS
147478
For mass defect of $0.3 \%$, the binding energy of $1 \mathrm{~kg}$ material is
1 $2.7 \times 10^{14} \mathrm{erg}$
2 $2.7 \times 10^{14} \mathrm{~J}$
3 $2.7 \times 10^{-14} \mathrm{erg}$
4 $2.7 \times 10^{-14} \mathrm{~J}$
Explanation:
B Given, Mass defect $(\Delta \mathrm{m})=0.3 \%$ of $1 \mathrm{~kg}=3 \times 10^{-3} \mathrm{~kg}$ Speed $(\mathrm{c})=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ The binding energy $(E)=(\Delta \mathrm{m}) \mathrm{c}^{2}$ $\mathrm{E}=3 \times 10^{-3}\left(3 \times 10^{8}\right)^{2}$ $\mathrm{E}=27 \times 10^{13} \mathrm{~J}$ $\mathrm{E}=2.7 \times 10^{14} \mathrm{~J}$
COMEDK 2016
NUCLEAR PHYSICS
147480
When an electron-positron pair annihilates, the energy released is about
1 $0.8 \times 10^{-13} \mathrm{~J}$
2 $1.6 \times 10^{-13} \mathrm{~J}$
3 $3.2 \times 10^{-13} \mathrm{~J}$
4 $4.8 \times 10^{-13} \mathrm{~J}$
Explanation:
B Rest mass of electron and positron $(\mathrm{m})=9.1$ $\times 10^{-31} \mathrm{~kg}$ During annihilation rest mass will convert into energy. Then, $\mathrm{E}=\Delta \mathrm{mc}^{2}$ $\mathrm{E}=2 \mathrm{mc}^{2}$ $\mathrm{E}=2 \times 9.1 \times 10^{-31} \times\left(3 \times 10^8\right)^2$ $\mathrm{E}=2 \times 9.1 \times 10^{-31} \times 9 \times 10^{16} \mathrm{~J}$ So, \(\quad E=1.638 \times 10^{-13} \mathrm{~J}\)
AIIMS-2004
NUCLEAR PHYSICS
147482
The binding energy per nucleon of ${ }^{16} \mathrm{O}$ is 7.97 $\mathrm{MeV}$ and that of ${ }^{17} \mathrm{O}$ is $7.75 \mathrm{MeV}$. The energy in $\mathrm{MeV}$ required to remove a neutron from ${ }^{17} \mathrm{O}$ is
1 3.52
2 3.64
3 4.23
4 7.86
5 1.68
Explanation:
C The binding energy per nucleon for $\mathrm{O}^{16}=$ $7.97 \mathrm{MeV}$ The binding energy per nucleon for $\mathrm{O}^{17}=7.75 \mathrm{MeV}$ Energy required $=$ Binding energy of $\mathrm{O}^{17}-$ Binding energy of $\mathrm{O}^{16}$ $=17 \times 7.75-16 \times 7.97$ $=131.75-127.52$ Energy required $=4.23 \mathrm{MeV}$
147477
Binding energy per nucleon versus 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
1 $\mathrm{Y} \rightarrow 2 \mathrm{Z}$
2 $\mathrm{W} \rightarrow \mathrm{X}+\mathrm{Z}$
3 $\mathrm{W} \rightarrow 2 \mathrm{Y}$
4 $\mathrm{X} \rightarrow \mathrm{Y}+\mathrm{Z}$
Explanation:
C Energy is released is a process when total binding energy of products is more than the reactants. In process- $\mathrm{W} \rightarrow 2 \mathrm{Y}$ Binding energy of reactants $W,\left|E_{W}\right|=120 \times 7.5=900$ $\mathrm{MeV}$ Binding energy of products $2 \mathrm{Y},\left|\mathrm{E}_{2 \mathrm{Y}}\right|=2(60 \times 8.5)=$ $1020 \mathrm{MeV}$ Here, $1020 \mathrm{MeV}>900 \mathrm{MeV}$ Hence, option (c) is correct.
AIIMS-2016
NUCLEAR PHYSICS
147478
For mass defect of $0.3 \%$, the binding energy of $1 \mathrm{~kg}$ material is
1 $2.7 \times 10^{14} \mathrm{erg}$
2 $2.7 \times 10^{14} \mathrm{~J}$
3 $2.7 \times 10^{-14} \mathrm{erg}$
4 $2.7 \times 10^{-14} \mathrm{~J}$
Explanation:
B Given, Mass defect $(\Delta \mathrm{m})=0.3 \%$ of $1 \mathrm{~kg}=3 \times 10^{-3} \mathrm{~kg}$ Speed $(\mathrm{c})=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ The binding energy $(E)=(\Delta \mathrm{m}) \mathrm{c}^{2}$ $\mathrm{E}=3 \times 10^{-3}\left(3 \times 10^{8}\right)^{2}$ $\mathrm{E}=27 \times 10^{13} \mathrm{~J}$ $\mathrm{E}=2.7 \times 10^{14} \mathrm{~J}$
COMEDK 2016
NUCLEAR PHYSICS
147480
When an electron-positron pair annihilates, the energy released is about
1 $0.8 \times 10^{-13} \mathrm{~J}$
2 $1.6 \times 10^{-13} \mathrm{~J}$
3 $3.2 \times 10^{-13} \mathrm{~J}$
4 $4.8 \times 10^{-13} \mathrm{~J}$
Explanation:
B Rest mass of electron and positron $(\mathrm{m})=9.1$ $\times 10^{-31} \mathrm{~kg}$ During annihilation rest mass will convert into energy. Then, $\mathrm{E}=\Delta \mathrm{mc}^{2}$ $\mathrm{E}=2 \mathrm{mc}^{2}$ $\mathrm{E}=2 \times 9.1 \times 10^{-31} \times\left(3 \times 10^8\right)^2$ $\mathrm{E}=2 \times 9.1 \times 10^{-31} \times 9 \times 10^{16} \mathrm{~J}$ So, \(\quad E=1.638 \times 10^{-13} \mathrm{~J}\)
AIIMS-2004
NUCLEAR PHYSICS
147482
The binding energy per nucleon of ${ }^{16} \mathrm{O}$ is 7.97 $\mathrm{MeV}$ and that of ${ }^{17} \mathrm{O}$ is $7.75 \mathrm{MeV}$. The energy in $\mathrm{MeV}$ required to remove a neutron from ${ }^{17} \mathrm{O}$ is
1 3.52
2 3.64
3 4.23
4 7.86
5 1.68
Explanation:
C The binding energy per nucleon for $\mathrm{O}^{16}=$ $7.97 \mathrm{MeV}$ The binding energy per nucleon for $\mathrm{O}^{17}=7.75 \mathrm{MeV}$ Energy required $=$ Binding energy of $\mathrm{O}^{17}-$ Binding energy of $\mathrm{O}^{16}$ $=17 \times 7.75-16 \times 7.97$ $=131.75-127.52$ Energy required $=4.23 \mathrm{MeV}$
147477
Binding energy per nucleon versus 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
1 $\mathrm{Y} \rightarrow 2 \mathrm{Z}$
2 $\mathrm{W} \rightarrow \mathrm{X}+\mathrm{Z}$
3 $\mathrm{W} \rightarrow 2 \mathrm{Y}$
4 $\mathrm{X} \rightarrow \mathrm{Y}+\mathrm{Z}$
Explanation:
C Energy is released is a process when total binding energy of products is more than the reactants. In process- $\mathrm{W} \rightarrow 2 \mathrm{Y}$ Binding energy of reactants $W,\left|E_{W}\right|=120 \times 7.5=900$ $\mathrm{MeV}$ Binding energy of products $2 \mathrm{Y},\left|\mathrm{E}_{2 \mathrm{Y}}\right|=2(60 \times 8.5)=$ $1020 \mathrm{MeV}$ Here, $1020 \mathrm{MeV}>900 \mathrm{MeV}$ Hence, option (c) is correct.
AIIMS-2016
NUCLEAR PHYSICS
147478
For mass defect of $0.3 \%$, the binding energy of $1 \mathrm{~kg}$ material is
1 $2.7 \times 10^{14} \mathrm{erg}$
2 $2.7 \times 10^{14} \mathrm{~J}$
3 $2.7 \times 10^{-14} \mathrm{erg}$
4 $2.7 \times 10^{-14} \mathrm{~J}$
Explanation:
B Given, Mass defect $(\Delta \mathrm{m})=0.3 \%$ of $1 \mathrm{~kg}=3 \times 10^{-3} \mathrm{~kg}$ Speed $(\mathrm{c})=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ The binding energy $(E)=(\Delta \mathrm{m}) \mathrm{c}^{2}$ $\mathrm{E}=3 \times 10^{-3}\left(3 \times 10^{8}\right)^{2}$ $\mathrm{E}=27 \times 10^{13} \mathrm{~J}$ $\mathrm{E}=2.7 \times 10^{14} \mathrm{~J}$
COMEDK 2016
NUCLEAR PHYSICS
147480
When an electron-positron pair annihilates, the energy released is about
1 $0.8 \times 10^{-13} \mathrm{~J}$
2 $1.6 \times 10^{-13} \mathrm{~J}$
3 $3.2 \times 10^{-13} \mathrm{~J}$
4 $4.8 \times 10^{-13} \mathrm{~J}$
Explanation:
B Rest mass of electron and positron $(\mathrm{m})=9.1$ $\times 10^{-31} \mathrm{~kg}$ During annihilation rest mass will convert into energy. Then, $\mathrm{E}=\Delta \mathrm{mc}^{2}$ $\mathrm{E}=2 \mathrm{mc}^{2}$ $\mathrm{E}=2 \times 9.1 \times 10^{-31} \times\left(3 \times 10^8\right)^2$ $\mathrm{E}=2 \times 9.1 \times 10^{-31} \times 9 \times 10^{16} \mathrm{~J}$ So, \(\quad E=1.638 \times 10^{-13} \mathrm{~J}\)
AIIMS-2004
NUCLEAR PHYSICS
147482
The binding energy per nucleon of ${ }^{16} \mathrm{O}$ is 7.97 $\mathrm{MeV}$ and that of ${ }^{17} \mathrm{O}$ is $7.75 \mathrm{MeV}$. The energy in $\mathrm{MeV}$ required to remove a neutron from ${ }^{17} \mathrm{O}$ is
1 3.52
2 3.64
3 4.23
4 7.86
5 1.68
Explanation:
C The binding energy per nucleon for $\mathrm{O}^{16}=$ $7.97 \mathrm{MeV}$ The binding energy per nucleon for $\mathrm{O}^{17}=7.75 \mathrm{MeV}$ Energy required $=$ Binding energy of $\mathrm{O}^{17}-$ Binding energy of $\mathrm{O}^{16}$ $=17 \times 7.75-16 \times 7.97$ $=131.75-127.52$ Energy required $=4.23 \mathrm{MeV}$
