148017
If the binding energy per nuclear in $\mathrm{Li}^{7}$ and $\mathrm{He}^{4}$ nuclei are respectively $5.60 \mathrm{Me} \mathrm{V}$ and 7.06 $\mathrm{MeV}$, then energy of reactor $\mathbf{L i}^{7}+\mathbf{P} \rightarrow 2_{2} \mathbf{H e}^{4} \text { is }$
1 $19.6 \mathrm{MeV}$
2 $2.4 \mathrm{MeV}$
3 $8.4 \mathrm{MeV}$
4 $17.3 \mathrm{MeV}$
Explanation:
D Given that, Binding energy of $\left(\mathrm{Li}^{7}\right)=7 \times 5.60 \mathrm{MeV}=39.20 \mathrm{MeV}$ And, binding energy of $\mathrm{He}^{4}=4 \times 7.06 \mathrm{MeV}=28.24$ $\mathrm{MeV}$ On applying principle of energy conversation energy of proton $=$ Total B.E. of $2 \times \mathrm{He}^{4}-$ energy of $\mathrm{Li}^{7}$ $=2 \times 28.24-39.20$ $=56.48-39.20$ $=17.28 \mathrm{MeV}$ $=17.3 \mathrm{MeV}$
VITEEE-2012
NUCLEAR PHYSICS
148022
On bombardment of $\mathrm{U}^{235}$ by slow neutrons, 200 $\mathrm{MeV}$ energy is released. If the power output of atomic reactor is $1.6 \mathrm{MW}$, then the rate of fission will be
1 $5 \times 10^{16}$ per second
2 $10 \times 10^{16}$ per second
3 $15 \times 10^{16}$ per second
4 $20 \times 10^{16}$ per second
Explanation:
A Given that Energy release per fission $(\mathrm{E})=200 \mathrm{MeV}$ Output power of the reactor $(\mathrm{P})=1.6 \mathrm{MW}$ $\therefore \mathrm{E}=200 \times 1.6 \times 10^{-19} \mathrm{MJ}$ $\because$ We know that $\text { Rate of fission }(\mathrm{R})=\frac{\mathrm{P}}{\mathrm{E}}$ $\therefore \quad \mathrm{R}=\frac{1.6}{200 \times 1.6 \times 10^{-19}}$ $\mathrm{R}=5 \times 10^{16} \text { per second }$
J and K CET- 2005
NUCLEAR PHYSICS
148018
Assume the graph of specific binding energy versus mass number is as shown in the figure. Using this graph, select the correct choice from the following
1 Fusion of two nuclei of mass number lying in the range of $100 \lt \mathrm{A} \lt 200$ will release energy
2 Fusion of two nuclei of mass number lying in the range of $51 \lt \mathrm{A} \lt 100$ will release energy
3 Fusion of two nuclei of mass number lying in the range of $1 \lt \mathrm{A} \lt 50$ will release energy
4 Fission of the nucleus of mass number lying in the range of $100 \lt \mathrm{A} \lt 200$ will release energy when broken into two fragments
Explanation:
B For fusion of two atoms of atomic mass number in the range $51 \lt \mathrm{A} \lt 100$ will combine and produce atom of the atomic number of range 100 to 150 and will be stable than individual atoms by releasing the energy.
Karnataka CET-2010
NUCLEAR PHYSICS
148024
In the fission of uranium, $0.5 \mathrm{~g}$ mass is decayed, then how much energy will be obtained by it?
1 1.25 kilo watt hour
2 $1.25 \times 10^{7}$ kilo watt hour
3 0.25 kilo watt hour
4 $1.25 \times 10^{4}$ kilo watt hour
Explanation:
B Given that, Mass of decayed $(\Delta \mathrm{m})=0.5 \mathrm{~g}=0.5 \times 10^{-3} \mathrm{~kg}$ Energy released $(\mathrm{E})=\Delta \mathrm{mc}^{2}$ $\mathrm{E}=0.5 \times 10^{-3} \times\left(3 \times 10^{8}\right)^{2}$ $\mathrm{E}=4.5 \times 10^{13} \mathrm{~J}$ $\mathrm{E}==\frac{4.5 \times 10^{13}}{3.6 \times 10^{6}}$ $\mathrm{E}=1.25 \times 10^{7} \mathrm{kWh}$
J and K CET- 2002
NUCLEAR PHYSICS
148026
In the nuclear fission reaction ${ }_{2}^{4} \mathrm{He}+{ }_{7}^{14} \mathrm{~N} \rightarrow{ }_{\mathrm{p}}^{\mathrm{q}} \mathrm{X}+{ }_{1}^{1} \mathrm{H}$ The nucleus ${ }_{\mathrm{p}}^{\mathrm{q}} \mathrm{X}$ is
1 nitrogen of mass 16
2 nitrogen of mass 17
3 oxygen of mass 16
4 oxygen of mass 17
Explanation:
D The given nuclear fission reaction, ${ }_{2}^{4} \mathrm{He}+{ }_{7}^{14} \mathrm{~N} \rightarrow{ }_{\mathrm{p}}^{9} \mathrm{X}+{ }_{1}^{1} \mathrm{H}$ In order that reaction holds true mass number and atomic number remains conserved. Mass number - $4+14=q+1$ $q=18-1$ $q=17$ Atomic number - $2+7=p+1$ $p=9-1$ $p=8$ Hence, unknown element is an isotope of oxygen of mass 17.
148017
If the binding energy per nuclear in $\mathrm{Li}^{7}$ and $\mathrm{He}^{4}$ nuclei are respectively $5.60 \mathrm{Me} \mathrm{V}$ and 7.06 $\mathrm{MeV}$, then energy of reactor $\mathbf{L i}^{7}+\mathbf{P} \rightarrow 2_{2} \mathbf{H e}^{4} \text { is }$
1 $19.6 \mathrm{MeV}$
2 $2.4 \mathrm{MeV}$
3 $8.4 \mathrm{MeV}$
4 $17.3 \mathrm{MeV}$
Explanation:
D Given that, Binding energy of $\left(\mathrm{Li}^{7}\right)=7 \times 5.60 \mathrm{MeV}=39.20 \mathrm{MeV}$ And, binding energy of $\mathrm{He}^{4}=4 \times 7.06 \mathrm{MeV}=28.24$ $\mathrm{MeV}$ On applying principle of energy conversation energy of proton $=$ Total B.E. of $2 \times \mathrm{He}^{4}-$ energy of $\mathrm{Li}^{7}$ $=2 \times 28.24-39.20$ $=56.48-39.20$ $=17.28 \mathrm{MeV}$ $=17.3 \mathrm{MeV}$
VITEEE-2012
NUCLEAR PHYSICS
148022
On bombardment of $\mathrm{U}^{235}$ by slow neutrons, 200 $\mathrm{MeV}$ energy is released. If the power output of atomic reactor is $1.6 \mathrm{MW}$, then the rate of fission will be
1 $5 \times 10^{16}$ per second
2 $10 \times 10^{16}$ per second
3 $15 \times 10^{16}$ per second
4 $20 \times 10^{16}$ per second
Explanation:
A Given that Energy release per fission $(\mathrm{E})=200 \mathrm{MeV}$ Output power of the reactor $(\mathrm{P})=1.6 \mathrm{MW}$ $\therefore \mathrm{E}=200 \times 1.6 \times 10^{-19} \mathrm{MJ}$ $\because$ We know that $\text { Rate of fission }(\mathrm{R})=\frac{\mathrm{P}}{\mathrm{E}}$ $\therefore \quad \mathrm{R}=\frac{1.6}{200 \times 1.6 \times 10^{-19}}$ $\mathrm{R}=5 \times 10^{16} \text { per second }$
J and K CET- 2005
NUCLEAR PHYSICS
148018
Assume the graph of specific binding energy versus mass number is as shown in the figure. Using this graph, select the correct choice from the following
1 Fusion of two nuclei of mass number lying in the range of $100 \lt \mathrm{A} \lt 200$ will release energy
2 Fusion of two nuclei of mass number lying in the range of $51 \lt \mathrm{A} \lt 100$ will release energy
3 Fusion of two nuclei of mass number lying in the range of $1 \lt \mathrm{A} \lt 50$ will release energy
4 Fission of the nucleus of mass number lying in the range of $100 \lt \mathrm{A} \lt 200$ will release energy when broken into two fragments
Explanation:
B For fusion of two atoms of atomic mass number in the range $51 \lt \mathrm{A} \lt 100$ will combine and produce atom of the atomic number of range 100 to 150 and will be stable than individual atoms by releasing the energy.
Karnataka CET-2010
NUCLEAR PHYSICS
148024
In the fission of uranium, $0.5 \mathrm{~g}$ mass is decayed, then how much energy will be obtained by it?
1 1.25 kilo watt hour
2 $1.25 \times 10^{7}$ kilo watt hour
3 0.25 kilo watt hour
4 $1.25 \times 10^{4}$ kilo watt hour
Explanation:
B Given that, Mass of decayed $(\Delta \mathrm{m})=0.5 \mathrm{~g}=0.5 \times 10^{-3} \mathrm{~kg}$ Energy released $(\mathrm{E})=\Delta \mathrm{mc}^{2}$ $\mathrm{E}=0.5 \times 10^{-3} \times\left(3 \times 10^{8}\right)^{2}$ $\mathrm{E}=4.5 \times 10^{13} \mathrm{~J}$ $\mathrm{E}==\frac{4.5 \times 10^{13}}{3.6 \times 10^{6}}$ $\mathrm{E}=1.25 \times 10^{7} \mathrm{kWh}$
J and K CET- 2002
NUCLEAR PHYSICS
148026
In the nuclear fission reaction ${ }_{2}^{4} \mathrm{He}+{ }_{7}^{14} \mathrm{~N} \rightarrow{ }_{\mathrm{p}}^{\mathrm{q}} \mathrm{X}+{ }_{1}^{1} \mathrm{H}$ The nucleus ${ }_{\mathrm{p}}^{\mathrm{q}} \mathrm{X}$ is
1 nitrogen of mass 16
2 nitrogen of mass 17
3 oxygen of mass 16
4 oxygen of mass 17
Explanation:
D The given nuclear fission reaction, ${ }_{2}^{4} \mathrm{He}+{ }_{7}^{14} \mathrm{~N} \rightarrow{ }_{\mathrm{p}}^{9} \mathrm{X}+{ }_{1}^{1} \mathrm{H}$ In order that reaction holds true mass number and atomic number remains conserved. Mass number - $4+14=q+1$ $q=18-1$ $q=17$ Atomic number - $2+7=p+1$ $p=9-1$ $p=8$ Hence, unknown element is an isotope of oxygen of mass 17.
148017
If the binding energy per nuclear in $\mathrm{Li}^{7}$ and $\mathrm{He}^{4}$ nuclei are respectively $5.60 \mathrm{Me} \mathrm{V}$ and 7.06 $\mathrm{MeV}$, then energy of reactor $\mathbf{L i}^{7}+\mathbf{P} \rightarrow 2_{2} \mathbf{H e}^{4} \text { is }$
1 $19.6 \mathrm{MeV}$
2 $2.4 \mathrm{MeV}$
3 $8.4 \mathrm{MeV}$
4 $17.3 \mathrm{MeV}$
Explanation:
D Given that, Binding energy of $\left(\mathrm{Li}^{7}\right)=7 \times 5.60 \mathrm{MeV}=39.20 \mathrm{MeV}$ And, binding energy of $\mathrm{He}^{4}=4 \times 7.06 \mathrm{MeV}=28.24$ $\mathrm{MeV}$ On applying principle of energy conversation energy of proton $=$ Total B.E. of $2 \times \mathrm{He}^{4}-$ energy of $\mathrm{Li}^{7}$ $=2 \times 28.24-39.20$ $=56.48-39.20$ $=17.28 \mathrm{MeV}$ $=17.3 \mathrm{MeV}$
VITEEE-2012
NUCLEAR PHYSICS
148022
On bombardment of $\mathrm{U}^{235}$ by slow neutrons, 200 $\mathrm{MeV}$ energy is released. If the power output of atomic reactor is $1.6 \mathrm{MW}$, then the rate of fission will be
1 $5 \times 10^{16}$ per second
2 $10 \times 10^{16}$ per second
3 $15 \times 10^{16}$ per second
4 $20 \times 10^{16}$ per second
Explanation:
A Given that Energy release per fission $(\mathrm{E})=200 \mathrm{MeV}$ Output power of the reactor $(\mathrm{P})=1.6 \mathrm{MW}$ $\therefore \mathrm{E}=200 \times 1.6 \times 10^{-19} \mathrm{MJ}$ $\because$ We know that $\text { Rate of fission }(\mathrm{R})=\frac{\mathrm{P}}{\mathrm{E}}$ $\therefore \quad \mathrm{R}=\frac{1.6}{200 \times 1.6 \times 10^{-19}}$ $\mathrm{R}=5 \times 10^{16} \text { per second }$
J and K CET- 2005
NUCLEAR PHYSICS
148018
Assume the graph of specific binding energy versus mass number is as shown in the figure. Using this graph, select the correct choice from the following
1 Fusion of two nuclei of mass number lying in the range of $100 \lt \mathrm{A} \lt 200$ will release energy
2 Fusion of two nuclei of mass number lying in the range of $51 \lt \mathrm{A} \lt 100$ will release energy
3 Fusion of two nuclei of mass number lying in the range of $1 \lt \mathrm{A} \lt 50$ will release energy
4 Fission of the nucleus of mass number lying in the range of $100 \lt \mathrm{A} \lt 200$ will release energy when broken into two fragments
Explanation:
B For fusion of two atoms of atomic mass number in the range $51 \lt \mathrm{A} \lt 100$ will combine and produce atom of the atomic number of range 100 to 150 and will be stable than individual atoms by releasing the energy.
Karnataka CET-2010
NUCLEAR PHYSICS
148024
In the fission of uranium, $0.5 \mathrm{~g}$ mass is decayed, then how much energy will be obtained by it?
1 1.25 kilo watt hour
2 $1.25 \times 10^{7}$ kilo watt hour
3 0.25 kilo watt hour
4 $1.25 \times 10^{4}$ kilo watt hour
Explanation:
B Given that, Mass of decayed $(\Delta \mathrm{m})=0.5 \mathrm{~g}=0.5 \times 10^{-3} \mathrm{~kg}$ Energy released $(\mathrm{E})=\Delta \mathrm{mc}^{2}$ $\mathrm{E}=0.5 \times 10^{-3} \times\left(3 \times 10^{8}\right)^{2}$ $\mathrm{E}=4.5 \times 10^{13} \mathrm{~J}$ $\mathrm{E}==\frac{4.5 \times 10^{13}}{3.6 \times 10^{6}}$ $\mathrm{E}=1.25 \times 10^{7} \mathrm{kWh}$
J and K CET- 2002
NUCLEAR PHYSICS
148026
In the nuclear fission reaction ${ }_{2}^{4} \mathrm{He}+{ }_{7}^{14} \mathrm{~N} \rightarrow{ }_{\mathrm{p}}^{\mathrm{q}} \mathrm{X}+{ }_{1}^{1} \mathrm{H}$ The nucleus ${ }_{\mathrm{p}}^{\mathrm{q}} \mathrm{X}$ is
1 nitrogen of mass 16
2 nitrogen of mass 17
3 oxygen of mass 16
4 oxygen of mass 17
Explanation:
D The given nuclear fission reaction, ${ }_{2}^{4} \mathrm{He}+{ }_{7}^{14} \mathrm{~N} \rightarrow{ }_{\mathrm{p}}^{9} \mathrm{X}+{ }_{1}^{1} \mathrm{H}$ In order that reaction holds true mass number and atomic number remains conserved. Mass number - $4+14=q+1$ $q=18-1$ $q=17$ Atomic number - $2+7=p+1$ $p=9-1$ $p=8$ Hence, unknown element is an isotope of oxygen of mass 17.
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
NUCLEAR PHYSICS
148017
If the binding energy per nuclear in $\mathrm{Li}^{7}$ and $\mathrm{He}^{4}$ nuclei are respectively $5.60 \mathrm{Me} \mathrm{V}$ and 7.06 $\mathrm{MeV}$, then energy of reactor $\mathbf{L i}^{7}+\mathbf{P} \rightarrow 2_{2} \mathbf{H e}^{4} \text { is }$
1 $19.6 \mathrm{MeV}$
2 $2.4 \mathrm{MeV}$
3 $8.4 \mathrm{MeV}$
4 $17.3 \mathrm{MeV}$
Explanation:
D Given that, Binding energy of $\left(\mathrm{Li}^{7}\right)=7 \times 5.60 \mathrm{MeV}=39.20 \mathrm{MeV}$ And, binding energy of $\mathrm{He}^{4}=4 \times 7.06 \mathrm{MeV}=28.24$ $\mathrm{MeV}$ On applying principle of energy conversation energy of proton $=$ Total B.E. of $2 \times \mathrm{He}^{4}-$ energy of $\mathrm{Li}^{7}$ $=2 \times 28.24-39.20$ $=56.48-39.20$ $=17.28 \mathrm{MeV}$ $=17.3 \mathrm{MeV}$
VITEEE-2012
NUCLEAR PHYSICS
148022
On bombardment of $\mathrm{U}^{235}$ by slow neutrons, 200 $\mathrm{MeV}$ energy is released. If the power output of atomic reactor is $1.6 \mathrm{MW}$, then the rate of fission will be
1 $5 \times 10^{16}$ per second
2 $10 \times 10^{16}$ per second
3 $15 \times 10^{16}$ per second
4 $20 \times 10^{16}$ per second
Explanation:
A Given that Energy release per fission $(\mathrm{E})=200 \mathrm{MeV}$ Output power of the reactor $(\mathrm{P})=1.6 \mathrm{MW}$ $\therefore \mathrm{E}=200 \times 1.6 \times 10^{-19} \mathrm{MJ}$ $\because$ We know that $\text { Rate of fission }(\mathrm{R})=\frac{\mathrm{P}}{\mathrm{E}}$ $\therefore \quad \mathrm{R}=\frac{1.6}{200 \times 1.6 \times 10^{-19}}$ $\mathrm{R}=5 \times 10^{16} \text { per second }$
J and K CET- 2005
NUCLEAR PHYSICS
148018
Assume the graph of specific binding energy versus mass number is as shown in the figure. Using this graph, select the correct choice from the following
1 Fusion of two nuclei of mass number lying in the range of $100 \lt \mathrm{A} \lt 200$ will release energy
2 Fusion of two nuclei of mass number lying in the range of $51 \lt \mathrm{A} \lt 100$ will release energy
3 Fusion of two nuclei of mass number lying in the range of $1 \lt \mathrm{A} \lt 50$ will release energy
4 Fission of the nucleus of mass number lying in the range of $100 \lt \mathrm{A} \lt 200$ will release energy when broken into two fragments
Explanation:
B For fusion of two atoms of atomic mass number in the range $51 \lt \mathrm{A} \lt 100$ will combine and produce atom of the atomic number of range 100 to 150 and will be stable than individual atoms by releasing the energy.
Karnataka CET-2010
NUCLEAR PHYSICS
148024
In the fission of uranium, $0.5 \mathrm{~g}$ mass is decayed, then how much energy will be obtained by it?
1 1.25 kilo watt hour
2 $1.25 \times 10^{7}$ kilo watt hour
3 0.25 kilo watt hour
4 $1.25 \times 10^{4}$ kilo watt hour
Explanation:
B Given that, Mass of decayed $(\Delta \mathrm{m})=0.5 \mathrm{~g}=0.5 \times 10^{-3} \mathrm{~kg}$ Energy released $(\mathrm{E})=\Delta \mathrm{mc}^{2}$ $\mathrm{E}=0.5 \times 10^{-3} \times\left(3 \times 10^{8}\right)^{2}$ $\mathrm{E}=4.5 \times 10^{13} \mathrm{~J}$ $\mathrm{E}==\frac{4.5 \times 10^{13}}{3.6 \times 10^{6}}$ $\mathrm{E}=1.25 \times 10^{7} \mathrm{kWh}$
J and K CET- 2002
NUCLEAR PHYSICS
148026
In the nuclear fission reaction ${ }_{2}^{4} \mathrm{He}+{ }_{7}^{14} \mathrm{~N} \rightarrow{ }_{\mathrm{p}}^{\mathrm{q}} \mathrm{X}+{ }_{1}^{1} \mathrm{H}$ The nucleus ${ }_{\mathrm{p}}^{\mathrm{q}} \mathrm{X}$ is
1 nitrogen of mass 16
2 nitrogen of mass 17
3 oxygen of mass 16
4 oxygen of mass 17
Explanation:
D The given nuclear fission reaction, ${ }_{2}^{4} \mathrm{He}+{ }_{7}^{14} \mathrm{~N} \rightarrow{ }_{\mathrm{p}}^{9} \mathrm{X}+{ }_{1}^{1} \mathrm{H}$ In order that reaction holds true mass number and atomic number remains conserved. Mass number - $4+14=q+1$ $q=18-1$ $q=17$ Atomic number - $2+7=p+1$ $p=9-1$ $p=8$ Hence, unknown element is an isotope of oxygen of mass 17.
148017
If the binding energy per nuclear in $\mathrm{Li}^{7}$ and $\mathrm{He}^{4}$ nuclei are respectively $5.60 \mathrm{Me} \mathrm{V}$ and 7.06 $\mathrm{MeV}$, then energy of reactor $\mathbf{L i}^{7}+\mathbf{P} \rightarrow 2_{2} \mathbf{H e}^{4} \text { is }$
1 $19.6 \mathrm{MeV}$
2 $2.4 \mathrm{MeV}$
3 $8.4 \mathrm{MeV}$
4 $17.3 \mathrm{MeV}$
Explanation:
D Given that, Binding energy of $\left(\mathrm{Li}^{7}\right)=7 \times 5.60 \mathrm{MeV}=39.20 \mathrm{MeV}$ And, binding energy of $\mathrm{He}^{4}=4 \times 7.06 \mathrm{MeV}=28.24$ $\mathrm{MeV}$ On applying principle of energy conversation energy of proton $=$ Total B.E. of $2 \times \mathrm{He}^{4}-$ energy of $\mathrm{Li}^{7}$ $=2 \times 28.24-39.20$ $=56.48-39.20$ $=17.28 \mathrm{MeV}$ $=17.3 \mathrm{MeV}$
VITEEE-2012
NUCLEAR PHYSICS
148022
On bombardment of $\mathrm{U}^{235}$ by slow neutrons, 200 $\mathrm{MeV}$ energy is released. If the power output of atomic reactor is $1.6 \mathrm{MW}$, then the rate of fission will be
1 $5 \times 10^{16}$ per second
2 $10 \times 10^{16}$ per second
3 $15 \times 10^{16}$ per second
4 $20 \times 10^{16}$ per second
Explanation:
A Given that Energy release per fission $(\mathrm{E})=200 \mathrm{MeV}$ Output power of the reactor $(\mathrm{P})=1.6 \mathrm{MW}$ $\therefore \mathrm{E}=200 \times 1.6 \times 10^{-19} \mathrm{MJ}$ $\because$ We know that $\text { Rate of fission }(\mathrm{R})=\frac{\mathrm{P}}{\mathrm{E}}$ $\therefore \quad \mathrm{R}=\frac{1.6}{200 \times 1.6 \times 10^{-19}}$ $\mathrm{R}=5 \times 10^{16} \text { per second }$
J and K CET- 2005
NUCLEAR PHYSICS
148018
Assume the graph of specific binding energy versus mass number is as shown in the figure. Using this graph, select the correct choice from the following
1 Fusion of two nuclei of mass number lying in the range of $100 \lt \mathrm{A} \lt 200$ will release energy
2 Fusion of two nuclei of mass number lying in the range of $51 \lt \mathrm{A} \lt 100$ will release energy
3 Fusion of two nuclei of mass number lying in the range of $1 \lt \mathrm{A} \lt 50$ will release energy
4 Fission of the nucleus of mass number lying in the range of $100 \lt \mathrm{A} \lt 200$ will release energy when broken into two fragments
Explanation:
B For fusion of two atoms of atomic mass number in the range $51 \lt \mathrm{A} \lt 100$ will combine and produce atom of the atomic number of range 100 to 150 and will be stable than individual atoms by releasing the energy.
Karnataka CET-2010
NUCLEAR PHYSICS
148024
In the fission of uranium, $0.5 \mathrm{~g}$ mass is decayed, then how much energy will be obtained by it?
1 1.25 kilo watt hour
2 $1.25 \times 10^{7}$ kilo watt hour
3 0.25 kilo watt hour
4 $1.25 \times 10^{4}$ kilo watt hour
Explanation:
B Given that, Mass of decayed $(\Delta \mathrm{m})=0.5 \mathrm{~g}=0.5 \times 10^{-3} \mathrm{~kg}$ Energy released $(\mathrm{E})=\Delta \mathrm{mc}^{2}$ $\mathrm{E}=0.5 \times 10^{-3} \times\left(3 \times 10^{8}\right)^{2}$ $\mathrm{E}=4.5 \times 10^{13} \mathrm{~J}$ $\mathrm{E}==\frac{4.5 \times 10^{13}}{3.6 \times 10^{6}}$ $\mathrm{E}=1.25 \times 10^{7} \mathrm{kWh}$
J and K CET- 2002
NUCLEAR PHYSICS
148026
In the nuclear fission reaction ${ }_{2}^{4} \mathrm{He}+{ }_{7}^{14} \mathrm{~N} \rightarrow{ }_{\mathrm{p}}^{\mathrm{q}} \mathrm{X}+{ }_{1}^{1} \mathrm{H}$ The nucleus ${ }_{\mathrm{p}}^{\mathrm{q}} \mathrm{X}$ is
1 nitrogen of mass 16
2 nitrogen of mass 17
3 oxygen of mass 16
4 oxygen of mass 17
Explanation:
D The given nuclear fission reaction, ${ }_{2}^{4} \mathrm{He}+{ }_{7}^{14} \mathrm{~N} \rightarrow{ }_{\mathrm{p}}^{9} \mathrm{X}+{ }_{1}^{1} \mathrm{H}$ In order that reaction holds true mass number and atomic number remains conserved. Mass number - $4+14=q+1$ $q=18-1$ $q=17$ Atomic number - $2+7=p+1$ $p=9-1$ $p=8$ Hence, unknown element is an isotope of oxygen of mass 17.
