2 Most of energy of fission is in the from of heat
3 In a fission of $\mathrm{U}^{235}$ about $200 \mathrm{eV}$ energy is released
4 On an average, one neutron is released per fission of $\mathrm{U}^{235}$
Explanation:
A The reaction, ${ }_{92} \mathrm{U}^{235}+{ }_{0} \mathrm{n}^{1} \rightarrow{ }_{92} \mathrm{U}^{236}$ ${ }_{92} \mathrm{U}^{236} \rightarrow{ }_{56} \mathrm{Ba}^{144}+{ }_{36} \mathrm{Kr}^{89}+3{ }_{0} \mathrm{n}^{1}+200 \mathrm{MeV}$ In a nuclear fission reaction, there is splitting of an atom of larger atomic weight into smaller fragments with release of some amount of energy is converted into energy. Hence, approx $0.1 \%$ mass is converted into energy.
CG PET- 2008
NUCLEAR PHYSICS
147956
The energy released per fission of a ${ }_{92} U^{235}$ nucleus is nearly-
1 $200 \mathrm{eV}$
2 $20 \mathrm{eV}$
3 $200 \mathrm{MeV}$
4 $2000 \mathrm{eV}$
Explanation:
C We know that, the energy released in $\mathrm{U}^{235}$ fission process is about to $0.8 \mathrm{MeV}$ per nucleon. Hence, the energy released per atom is about $=0.8 \times 235$ $=188 \mathrm{MeV}$ $\approx 200 \mathrm{MeV}$
CG PET-22.05.2022
NUCLEAR PHYSICS
147962
If the energy of a hydrogen atom in $\mathbf{n}^{\text {th }}$ orbit is $E_{n}$ then energy in the $n^{\text {th }}$ orbit of a singly ionized helium atom will be
1 $4 \mathrm{E}_{\mathrm{n}}$
2 $\mathrm{E}_{\mathrm{n}} / 2$
3 $2 \mathrm{E}_{\mathrm{n}}$
4 $250 \mathrm{~m} / \mathrm{s}$
Explanation:
A The energy of a hydrogen atom in $n^{\text {th }}$ order $\mathrm{E}_{\mathrm{n}}=\frac{-13.6 \mathrm{Z}^{2}}{\mathrm{n}^{2}}$ $\frac{\mathrm{E}_{\mathrm{H}}}{\mathrm{E}_{\mathrm{He}}}=\left(\frac{\mathrm{Z}_{\mathrm{H}}}{\mathrm{Z}_{\mathrm{He}}}\right)^{2}$ $\frac{\mathrm{E}_{\mathrm{H}}}{\mathrm{E}_{\mathrm{He}}}=\left(\frac{1}{2}\right)^{2} \quad\left[\therefore \mathrm{E}_{\mathrm{H}}=\mathrm{E}_{\mathrm{n}}\right]$ $\mathrm{E}_{\mathrm{He}}=4 \mathrm{E}_{\mathrm{n}}$
Manipal UGET-2013
NUCLEAR PHYSICS
147964
On bombarding $\mathrm{U}^{235}$ by slow neutron, $200 \mathrm{MeV}$ energy is released. If the power output of atomic reactor is $1.6 \mathrm{MW}$, then the rate of fission will be :
1 $5 \times 10^{22} / \mathrm{s}$
2 $5 \times 10^{16} / \mathrm{s}$
3 $8 \times 10^{16} / \mathrm{s}$
4 $20 \times 10^{16} / \mathrm{s}$
Explanation:
B Given, Output power of the reactor $(\mathrm{P})=1.6 \mathrm{MW}$ Energy released per fission $(\mathrm{E})=200 \mathrm{MeV}=200 \times 1.6$ $\times 10^{-19} \mathrm{MJ}$ So, rate of fission $(R)=\frac{P}{E}$ $\mathrm{R}=\frac{1.6}{200 \times 1.6 \times 10^{-19}}=5 \times 10^{16} / \mathrm{s} \text {. }$
2 Most of energy of fission is in the from of heat
3 In a fission of $\mathrm{U}^{235}$ about $200 \mathrm{eV}$ energy is released
4 On an average, one neutron is released per fission of $\mathrm{U}^{235}$
Explanation:
A The reaction, ${ }_{92} \mathrm{U}^{235}+{ }_{0} \mathrm{n}^{1} \rightarrow{ }_{92} \mathrm{U}^{236}$ ${ }_{92} \mathrm{U}^{236} \rightarrow{ }_{56} \mathrm{Ba}^{144}+{ }_{36} \mathrm{Kr}^{89}+3{ }_{0} \mathrm{n}^{1}+200 \mathrm{MeV}$ In a nuclear fission reaction, there is splitting of an atom of larger atomic weight into smaller fragments with release of some amount of energy is converted into energy. Hence, approx $0.1 \%$ mass is converted into energy.
CG PET- 2008
NUCLEAR PHYSICS
147956
The energy released per fission of a ${ }_{92} U^{235}$ nucleus is nearly-
1 $200 \mathrm{eV}$
2 $20 \mathrm{eV}$
3 $200 \mathrm{MeV}$
4 $2000 \mathrm{eV}$
Explanation:
C We know that, the energy released in $\mathrm{U}^{235}$ fission process is about to $0.8 \mathrm{MeV}$ per nucleon. Hence, the energy released per atom is about $=0.8 \times 235$ $=188 \mathrm{MeV}$ $\approx 200 \mathrm{MeV}$
CG PET-22.05.2022
NUCLEAR PHYSICS
147962
If the energy of a hydrogen atom in $\mathbf{n}^{\text {th }}$ orbit is $E_{n}$ then energy in the $n^{\text {th }}$ orbit of a singly ionized helium atom will be
1 $4 \mathrm{E}_{\mathrm{n}}$
2 $\mathrm{E}_{\mathrm{n}} / 2$
3 $2 \mathrm{E}_{\mathrm{n}}$
4 $250 \mathrm{~m} / \mathrm{s}$
Explanation:
A The energy of a hydrogen atom in $n^{\text {th }}$ order $\mathrm{E}_{\mathrm{n}}=\frac{-13.6 \mathrm{Z}^{2}}{\mathrm{n}^{2}}$ $\frac{\mathrm{E}_{\mathrm{H}}}{\mathrm{E}_{\mathrm{He}}}=\left(\frac{\mathrm{Z}_{\mathrm{H}}}{\mathrm{Z}_{\mathrm{He}}}\right)^{2}$ $\frac{\mathrm{E}_{\mathrm{H}}}{\mathrm{E}_{\mathrm{He}}}=\left(\frac{1}{2}\right)^{2} \quad\left[\therefore \mathrm{E}_{\mathrm{H}}=\mathrm{E}_{\mathrm{n}}\right]$ $\mathrm{E}_{\mathrm{He}}=4 \mathrm{E}_{\mathrm{n}}$
Manipal UGET-2013
NUCLEAR PHYSICS
147964
On bombarding $\mathrm{U}^{235}$ by slow neutron, $200 \mathrm{MeV}$ energy is released. If the power output of atomic reactor is $1.6 \mathrm{MW}$, then the rate of fission will be :
1 $5 \times 10^{22} / \mathrm{s}$
2 $5 \times 10^{16} / \mathrm{s}$
3 $8 \times 10^{16} / \mathrm{s}$
4 $20 \times 10^{16} / \mathrm{s}$
Explanation:
B Given, Output power of the reactor $(\mathrm{P})=1.6 \mathrm{MW}$ Energy released per fission $(\mathrm{E})=200 \mathrm{MeV}=200 \times 1.6$ $\times 10^{-19} \mathrm{MJ}$ So, rate of fission $(R)=\frac{P}{E}$ $\mathrm{R}=\frac{1.6}{200 \times 1.6 \times 10^{-19}}=5 \times 10^{16} / \mathrm{s} \text {. }$
2 Most of energy of fission is in the from of heat
3 In a fission of $\mathrm{U}^{235}$ about $200 \mathrm{eV}$ energy is released
4 On an average, one neutron is released per fission of $\mathrm{U}^{235}$
Explanation:
A The reaction, ${ }_{92} \mathrm{U}^{235}+{ }_{0} \mathrm{n}^{1} \rightarrow{ }_{92} \mathrm{U}^{236}$ ${ }_{92} \mathrm{U}^{236} \rightarrow{ }_{56} \mathrm{Ba}^{144}+{ }_{36} \mathrm{Kr}^{89}+3{ }_{0} \mathrm{n}^{1}+200 \mathrm{MeV}$ In a nuclear fission reaction, there is splitting of an atom of larger atomic weight into smaller fragments with release of some amount of energy is converted into energy. Hence, approx $0.1 \%$ mass is converted into energy.
CG PET- 2008
NUCLEAR PHYSICS
147956
The energy released per fission of a ${ }_{92} U^{235}$ nucleus is nearly-
1 $200 \mathrm{eV}$
2 $20 \mathrm{eV}$
3 $200 \mathrm{MeV}$
4 $2000 \mathrm{eV}$
Explanation:
C We know that, the energy released in $\mathrm{U}^{235}$ fission process is about to $0.8 \mathrm{MeV}$ per nucleon. Hence, the energy released per atom is about $=0.8 \times 235$ $=188 \mathrm{MeV}$ $\approx 200 \mathrm{MeV}$
CG PET-22.05.2022
NUCLEAR PHYSICS
147962
If the energy of a hydrogen atom in $\mathbf{n}^{\text {th }}$ orbit is $E_{n}$ then energy in the $n^{\text {th }}$ orbit of a singly ionized helium atom will be
1 $4 \mathrm{E}_{\mathrm{n}}$
2 $\mathrm{E}_{\mathrm{n}} / 2$
3 $2 \mathrm{E}_{\mathrm{n}}$
4 $250 \mathrm{~m} / \mathrm{s}$
Explanation:
A The energy of a hydrogen atom in $n^{\text {th }}$ order $\mathrm{E}_{\mathrm{n}}=\frac{-13.6 \mathrm{Z}^{2}}{\mathrm{n}^{2}}$ $\frac{\mathrm{E}_{\mathrm{H}}}{\mathrm{E}_{\mathrm{He}}}=\left(\frac{\mathrm{Z}_{\mathrm{H}}}{\mathrm{Z}_{\mathrm{He}}}\right)^{2}$ $\frac{\mathrm{E}_{\mathrm{H}}}{\mathrm{E}_{\mathrm{He}}}=\left(\frac{1}{2}\right)^{2} \quad\left[\therefore \mathrm{E}_{\mathrm{H}}=\mathrm{E}_{\mathrm{n}}\right]$ $\mathrm{E}_{\mathrm{He}}=4 \mathrm{E}_{\mathrm{n}}$
Manipal UGET-2013
NUCLEAR PHYSICS
147964
On bombarding $\mathrm{U}^{235}$ by slow neutron, $200 \mathrm{MeV}$ energy is released. If the power output of atomic reactor is $1.6 \mathrm{MW}$, then the rate of fission will be :
1 $5 \times 10^{22} / \mathrm{s}$
2 $5 \times 10^{16} / \mathrm{s}$
3 $8 \times 10^{16} / \mathrm{s}$
4 $20 \times 10^{16} / \mathrm{s}$
Explanation:
B Given, Output power of the reactor $(\mathrm{P})=1.6 \mathrm{MW}$ Energy released per fission $(\mathrm{E})=200 \mathrm{MeV}=200 \times 1.6$ $\times 10^{-19} \mathrm{MJ}$ So, rate of fission $(R)=\frac{P}{E}$ $\mathrm{R}=\frac{1.6}{200 \times 1.6 \times 10^{-19}}=5 \times 10^{16} / \mathrm{s} \text {. }$
2 Most of energy of fission is in the from of heat
3 In a fission of $\mathrm{U}^{235}$ about $200 \mathrm{eV}$ energy is released
4 On an average, one neutron is released per fission of $\mathrm{U}^{235}$
Explanation:
A The reaction, ${ }_{92} \mathrm{U}^{235}+{ }_{0} \mathrm{n}^{1} \rightarrow{ }_{92} \mathrm{U}^{236}$ ${ }_{92} \mathrm{U}^{236} \rightarrow{ }_{56} \mathrm{Ba}^{144}+{ }_{36} \mathrm{Kr}^{89}+3{ }_{0} \mathrm{n}^{1}+200 \mathrm{MeV}$ In a nuclear fission reaction, there is splitting of an atom of larger atomic weight into smaller fragments with release of some amount of energy is converted into energy. Hence, approx $0.1 \%$ mass is converted into energy.
CG PET- 2008
NUCLEAR PHYSICS
147956
The energy released per fission of a ${ }_{92} U^{235}$ nucleus is nearly-
1 $200 \mathrm{eV}$
2 $20 \mathrm{eV}$
3 $200 \mathrm{MeV}$
4 $2000 \mathrm{eV}$
Explanation:
C We know that, the energy released in $\mathrm{U}^{235}$ fission process is about to $0.8 \mathrm{MeV}$ per nucleon. Hence, the energy released per atom is about $=0.8 \times 235$ $=188 \mathrm{MeV}$ $\approx 200 \mathrm{MeV}$
CG PET-22.05.2022
NUCLEAR PHYSICS
147962
If the energy of a hydrogen atom in $\mathbf{n}^{\text {th }}$ orbit is $E_{n}$ then energy in the $n^{\text {th }}$ orbit of a singly ionized helium atom will be
1 $4 \mathrm{E}_{\mathrm{n}}$
2 $\mathrm{E}_{\mathrm{n}} / 2$
3 $2 \mathrm{E}_{\mathrm{n}}$
4 $250 \mathrm{~m} / \mathrm{s}$
Explanation:
A The energy of a hydrogen atom in $n^{\text {th }}$ order $\mathrm{E}_{\mathrm{n}}=\frac{-13.6 \mathrm{Z}^{2}}{\mathrm{n}^{2}}$ $\frac{\mathrm{E}_{\mathrm{H}}}{\mathrm{E}_{\mathrm{He}}}=\left(\frac{\mathrm{Z}_{\mathrm{H}}}{\mathrm{Z}_{\mathrm{He}}}\right)^{2}$ $\frac{\mathrm{E}_{\mathrm{H}}}{\mathrm{E}_{\mathrm{He}}}=\left(\frac{1}{2}\right)^{2} \quad\left[\therefore \mathrm{E}_{\mathrm{H}}=\mathrm{E}_{\mathrm{n}}\right]$ $\mathrm{E}_{\mathrm{He}}=4 \mathrm{E}_{\mathrm{n}}$
Manipal UGET-2013
NUCLEAR PHYSICS
147964
On bombarding $\mathrm{U}^{235}$ by slow neutron, $200 \mathrm{MeV}$ energy is released. If the power output of atomic reactor is $1.6 \mathrm{MW}$, then the rate of fission will be :
1 $5 \times 10^{22} / \mathrm{s}$
2 $5 \times 10^{16} / \mathrm{s}$
3 $8 \times 10^{16} / \mathrm{s}$
4 $20 \times 10^{16} / \mathrm{s}$
Explanation:
B Given, Output power of the reactor $(\mathrm{P})=1.6 \mathrm{MW}$ Energy released per fission $(\mathrm{E})=200 \mathrm{MeV}=200 \times 1.6$ $\times 10^{-19} \mathrm{MJ}$ So, rate of fission $(R)=\frac{P}{E}$ $\mathrm{R}=\frac{1.6}{200 \times 1.6 \times 10^{-19}}=5 \times 10^{16} / \mathrm{s} \text {. }$
