QBManager

NUCLEAR PHYSICS

147978 A $2 MeV$ neutron is emitted in a fission reactor. If it losses half of its kinetic energy in each collision with a moderator atom, how many collisions must it undergo to achieve thermal energy of $0.039 eV$ ?

1 20

2 26

3 30

4 42

5 48

Explanation:

B $2 MeV$ neutron losses half of its kinetic energy in each collision so It from a geometric progression of each collision.
$2 MeV, 1 MeV 0.5 MeV + \dots \dots \dots .0 .039 eV$
first terms $a = 2 MeV$ ratio $r = \frac{1}{2}$
$n^{th}$ term of G.P
$an = {ar}^{n - 1}$
$∴ 0.039 = 2 \times 10^{6} {(\frac{1}{2})}^{n - 1}$
$1.95 \times 10^{- 8} = {(\frac{1}{2})}^{n - 1}$
Taking $\log$ of both sides
$(n - 1) \log (\frac{1}{2}) = \log (1.95 \times 10^{- 8}$
$n = 26.6$
$n = 26$

Kerala CEE -2018

NUCLEAR PHYSICS

147982 Assertion: Energy is released when heavy nuclei undergo fission or light nuclei undergo fusion.
Reason: For heavy nuclei, binding energy per nucleon increases with increasing $Z$ while for light nuclei it decreases with increasing $Z$ .

1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion

3 If Assertion is correct but Reason is incorrect

4 If both the Assertion and Reason are incorrect

Explanation:

C We know that,
Energy is released when heavy nuclei undergo fission and lighter nuclei undergoes the nuclear fusion.
Hence, assertion is true.
For heavy nuclei the binding energy per nucleon decreases with increasing $z$ and for light nuclei, Binding energy per nucleon increases with increasing $Z$ .
So, reason is false.

AIIMS-2017

NUCLEAR PHYSICS

147983 When $_{92} U^{235}$ undergoes fission, $0.1 %$ of its original mass is changed into energy. How much energy is released if $1 kg$ of $_{92} U^{235}$ undergoes fission?

1

9 \times 10^{10} J

2

9 \times 10^{11} J

3

9 \times 10^{12} J

4

9 \times 10^{13} J

Explanation:

D Mass of Uranium $= 0.1 %$ of $1 kg$
$= \frac{0.1}{100} \times 1 = 10^{- 3} kg$
We know that,
$E = {mc}^{2}$
$E = 10^{- 3} \times {(3 \times 10^{8})}^{2}$
$E = 10^{- 3} \times 9 \times 10^{16}$
$E = 9 \times 10^{13} J$

BITSAT- 2017

NUCLEAR PHYSICS

147984 In a nuclear reactor, $U^{235}$ undergoes fission liberating $200 MeV$ of energy. The reactor has a $10 %$ efficiency and produces 1000 MW power. If the reactor is to function for $10 yr$ , then find the total mass of uranium required.

1

36.5 \times 10^{3} kg

2

36 \times 10^{3} kg

3

39.5 \times 10^{3} kg

4

38.2 \times 10^{3} kg

Explanation:

D Given that,
Energy liberating by fission $= 200 MeV$
Power produced in reactor $= 1000 MW$
Efficiency of reactor $= 10 %$
Total time of reactor to function $= 10$ years
Energy which reactor would supply for 10 years-
$E = 1000 \times 10^{6} \times (10 \times 365 \times 24 \times 3600)$
$E = 10^{9} \times (10 \times 365 \times 24 \times 3600)$
$E = 3.15 \times 10^{17} J$
When efficiency is $10 %$ the energy will be-
$η = \frac{Energy output}{Energy input}$
$E_{Input} = \frac{3.15 \times 10^{17}}{\frac{10}{100}} = 3.15 \times 10^{18} J$
$∵ E_{fission} = 200 \times 10^{6} \times 1.6 \times 10^{- 19}$
$E_{fission} = 3.2 \times 10^{- 11} J$
We know that,
$Total number of fission = \frac{E_{In}}{E_{fission}}$
$= \frac{3.15 \times 10^{18}}{3.2 \times 10^{- 11}}$
$= 9.8 \times 10^{28}$
We know that,
$Moles (n) = \frac{Total number of fission}{N_{A}}$
$n = \frac{9.8 \times 10^{28}}{6.02 \times 10^{26}}$
$n = 163.7$
Mass of uranium required is given by-
$m = 163.7 \times 235$
$m = 38.2 \times 10^{3} kg$

Manipal UGET-2015

NUCLEAR PHYSICS

147978 A $2 MeV$ neutron is emitted in a fission reactor. If it losses half of its kinetic energy in each collision with a moderator atom, how many collisions must it undergo to achieve thermal energy of $0.039 eV$ ?

1 20

2 26

3 30

4 42

5 48

Explanation:

B $2 MeV$ neutron losses half of its kinetic energy in each collision so It from a geometric progression of each collision.
$2 MeV, 1 MeV 0.5 MeV + \dots \dots \dots .0 .039 eV$
first terms $a = 2 MeV$ ratio $r = \frac{1}{2}$
$n^{th}$ term of G.P
$an = {ar}^{n - 1}$
$∴ 0.039 = 2 \times 10^{6} {(\frac{1}{2})}^{n - 1}$
$1.95 \times 10^{- 8} = {(\frac{1}{2})}^{n - 1}$
Taking $\log$ of both sides
$(n - 1) \log (\frac{1}{2}) = \log (1.95 \times 10^{- 8}$
$n = 26.6$
$n = 26$

Kerala CEE -2018

NUCLEAR PHYSICS

147982 Assertion: Energy is released when heavy nuclei undergo fission or light nuclei undergo fusion.
Reason: For heavy nuclei, binding energy per nucleon increases with increasing $Z$ while for light nuclei it decreases with increasing $Z$ .

1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion

3 If Assertion is correct but Reason is incorrect

4 If both the Assertion and Reason are incorrect

Explanation:

C We know that,
Energy is released when heavy nuclei undergo fission and lighter nuclei undergoes the nuclear fusion.
Hence, assertion is true.
For heavy nuclei the binding energy per nucleon decreases with increasing $z$ and for light nuclei, Binding energy per nucleon increases with increasing $Z$ .
So, reason is false.

AIIMS-2017

NUCLEAR PHYSICS

147983 When $_{92} U^{235}$ undergoes fission, $0.1 %$ of its original mass is changed into energy. How much energy is released if $1 kg$ of $_{92} U^{235}$ undergoes fission?

1

9 \times 10^{10} J

2

9 \times 10^{11} J

3

9 \times 10^{12} J

4

9 \times 10^{13} J

Explanation:

D Mass of Uranium $= 0.1 %$ of $1 kg$
$= \frac{0.1}{100} \times 1 = 10^{- 3} kg$
We know that,
$E = {mc}^{2}$
$E = 10^{- 3} \times {(3 \times 10^{8})}^{2}$
$E = 10^{- 3} \times 9 \times 10^{16}$
$E = 9 \times 10^{13} J$

BITSAT- 2017

NUCLEAR PHYSICS

147984 In a nuclear reactor, $U^{235}$ undergoes fission liberating $200 MeV$ of energy. The reactor has a $10 %$ efficiency and produces 1000 MW power. If the reactor is to function for $10 yr$ , then find the total mass of uranium required.

1

36.5 \times 10^{3} kg

2

36 \times 10^{3} kg

3

39.5 \times 10^{3} kg

4

38.2 \times 10^{3} kg

Explanation:

D Given that,
Energy liberating by fission $= 200 MeV$
Power produced in reactor $= 1000 MW$
Efficiency of reactor $= 10 %$
Total time of reactor to function $= 10$ years
Energy which reactor would supply for 10 years-
$E = 1000 \times 10^{6} \times (10 \times 365 \times 24 \times 3600)$
$E = 10^{9} \times (10 \times 365 \times 24 \times 3600)$
$E = 3.15 \times 10^{17} J$
When efficiency is $10 %$ the energy will be-
$η = \frac{Energy output}{Energy input}$
$E_{Input} = \frac{3.15 \times 10^{17}}{\frac{10}{100}} = 3.15 \times 10^{18} J$
$∵ E_{fission} = 200 \times 10^{6} \times 1.6 \times 10^{- 19}$
$E_{fission} = 3.2 \times 10^{- 11} J$
We know that,
$Total number of fission = \frac{E_{In}}{E_{fission}}$
$= \frac{3.15 \times 10^{18}}{3.2 \times 10^{- 11}}$
$= 9.8 \times 10^{28}$
We know that,
$Moles (n) = \frac{Total number of fission}{N_{A}}$
$n = \frac{9.8 \times 10^{28}}{6.02 \times 10^{26}}$
$n = 163.7$
Mass of uranium required is given by-
$m = 163.7 \times 235$
$m = 38.2 \times 10^{3} kg$

Manipal UGET-2015

NUCLEAR PHYSICS

147978 A $2 MeV$ neutron is emitted in a fission reactor. If it losses half of its kinetic energy in each collision with a moderator atom, how many collisions must it undergo to achieve thermal energy of $0.039 eV$ ?

1 20

2 26

3 30

4 42

5 48

Explanation:

B $2 MeV$ neutron losses half of its kinetic energy in each collision so It from a geometric progression of each collision.
$2 MeV, 1 MeV 0.5 MeV + \dots \dots \dots .0 .039 eV$
first terms $a = 2 MeV$ ratio $r = \frac{1}{2}$
$n^{th}$ term of G.P
$an = {ar}^{n - 1}$
$∴ 0.039 = 2 \times 10^{6} {(\frac{1}{2})}^{n - 1}$
$1.95 \times 10^{- 8} = {(\frac{1}{2})}^{n - 1}$
Taking $\log$ of both sides
$(n - 1) \log (\frac{1}{2}) = \log (1.95 \times 10^{- 8}$
$n = 26.6$
$n = 26$

Kerala CEE -2018

NUCLEAR PHYSICS

147982 Assertion: Energy is released when heavy nuclei undergo fission or light nuclei undergo fusion.
Reason: For heavy nuclei, binding energy per nucleon increases with increasing $Z$ while for light nuclei it decreases with increasing $Z$ .

1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion

3 If Assertion is correct but Reason is incorrect

4 If both the Assertion and Reason are incorrect

Explanation:

C We know that,
Energy is released when heavy nuclei undergo fission and lighter nuclei undergoes the nuclear fusion.
Hence, assertion is true.
For heavy nuclei the binding energy per nucleon decreases with increasing $z$ and for light nuclei, Binding energy per nucleon increases with increasing $Z$ .
So, reason is false.

AIIMS-2017

NUCLEAR PHYSICS

147983 When $_{92} U^{235}$ undergoes fission, $0.1 %$ of its original mass is changed into energy. How much energy is released if $1 kg$ of $_{92} U^{235}$ undergoes fission?

1

9 \times 10^{10} J

2

9 \times 10^{11} J

3

9 \times 10^{12} J

4

9 \times 10^{13} J

Explanation:

D Mass of Uranium $= 0.1 %$ of $1 kg$
$= \frac{0.1}{100} \times 1 = 10^{- 3} kg$
We know that,
$E = {mc}^{2}$
$E = 10^{- 3} \times {(3 \times 10^{8})}^{2}$
$E = 10^{- 3} \times 9 \times 10^{16}$
$E = 9 \times 10^{13} J$

BITSAT- 2017

NUCLEAR PHYSICS

147984 In a nuclear reactor, $U^{235}$ undergoes fission liberating $200 MeV$ of energy. The reactor has a $10 %$ efficiency and produces 1000 MW power. If the reactor is to function for $10 yr$ , then find the total mass of uranium required.

1

36.5 \times 10^{3} kg

2

36 \times 10^{3} kg

3

39.5 \times 10^{3} kg

4

38.2 \times 10^{3} kg

Explanation:

D Given that,
Energy liberating by fission $= 200 MeV$
Power produced in reactor $= 1000 MW$
Efficiency of reactor $= 10 %$
Total time of reactor to function $= 10$ years
Energy which reactor would supply for 10 years-
$E = 1000 \times 10^{6} \times (10 \times 365 \times 24 \times 3600)$
$E = 10^{9} \times (10 \times 365 \times 24 \times 3600)$
$E = 3.15 \times 10^{17} J$
When efficiency is $10 %$ the energy will be-
$η = \frac{Energy output}{Energy input}$
$E_{Input} = \frac{3.15 \times 10^{17}}{\frac{10}{100}} = 3.15 \times 10^{18} J$
$∵ E_{fission} = 200 \times 10^{6} \times 1.6 \times 10^{- 19}$
$E_{fission} = 3.2 \times 10^{- 11} J$
We know that,
$Total number of fission = \frac{E_{In}}{E_{fission}}$
$= \frac{3.15 \times 10^{18}}{3.2 \times 10^{- 11}}$
$= 9.8 \times 10^{28}$
We know that,
$Moles (n) = \frac{Total number of fission}{N_{A}}$
$n = \frac{9.8 \times 10^{28}}{6.02 \times 10^{26}}$
$n = 163.7$
Mass of uranium required is given by-
$m = 163.7 \times 235$
$m = 38.2 \times 10^{3} kg$

Manipal UGET-2015

NUCLEAR PHYSICS

147978 A $2 MeV$ neutron is emitted in a fission reactor. If it losses half of its kinetic energy in each collision with a moderator atom, how many collisions must it undergo to achieve thermal energy of $0.039 eV$ ?

1 20

2 26

3 30

4 42

5 48

Explanation:

B $2 MeV$ neutron losses half of its kinetic energy in each collision so It from a geometric progression of each collision.
$2 MeV, 1 MeV 0.5 MeV + \dots \dots \dots .0 .039 eV$
first terms $a = 2 MeV$ ratio $r = \frac{1}{2}$
$n^{th}$ term of G.P
$an = {ar}^{n - 1}$
$∴ 0.039 = 2 \times 10^{6} {(\frac{1}{2})}^{n - 1}$
$1.95 \times 10^{- 8} = {(\frac{1}{2})}^{n - 1}$
Taking $\log$ of both sides
$(n - 1) \log (\frac{1}{2}) = \log (1.95 \times 10^{- 8}$
$n = 26.6$
$n = 26$

Kerala CEE -2018

NUCLEAR PHYSICS

147982 Assertion: Energy is released when heavy nuclei undergo fission or light nuclei undergo fusion.
Reason: For heavy nuclei, binding energy per nucleon increases with increasing $Z$ while for light nuclei it decreases with increasing $Z$ .

1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion

3 If Assertion is correct but Reason is incorrect

4 If both the Assertion and Reason are incorrect

Explanation:

C We know that,
Energy is released when heavy nuclei undergo fission and lighter nuclei undergoes the nuclear fusion.
Hence, assertion is true.
For heavy nuclei the binding energy per nucleon decreases with increasing $z$ and for light nuclei, Binding energy per nucleon increases with increasing $Z$ .
So, reason is false.

AIIMS-2017

NUCLEAR PHYSICS

147983 When $_{92} U^{235}$ undergoes fission, $0.1 %$ of its original mass is changed into energy. How much energy is released if $1 kg$ of $_{92} U^{235}$ undergoes fission?

1

9 \times 10^{10} J

2

9 \times 10^{11} J

3

9 \times 10^{12} J

4

9 \times 10^{13} J

Explanation:

D Mass of Uranium $= 0.1 %$ of $1 kg$
$= \frac{0.1}{100} \times 1 = 10^{- 3} kg$
We know that,
$E = {mc}^{2}$
$E = 10^{- 3} \times {(3 \times 10^{8})}^{2}$
$E = 10^{- 3} \times 9 \times 10^{16}$
$E = 9 \times 10^{13} J$

BITSAT- 2017

NUCLEAR PHYSICS

147984 In a nuclear reactor, $U^{235}$ undergoes fission liberating $200 MeV$ of energy. The reactor has a $10 %$ efficiency and produces 1000 MW power. If the reactor is to function for $10 yr$ , then find the total mass of uranium required.

1

36.5 \times 10^{3} kg

2

36 \times 10^{3} kg

3

39.5 \times 10^{3} kg

4

38.2 \times 10^{3} kg

Explanation:

D Given that,
Energy liberating by fission $= 200 MeV$
Power produced in reactor $= 1000 MW$
Efficiency of reactor $= 10 %$
Total time of reactor to function $= 10$ years
Energy which reactor would supply for 10 years-
$E = 1000 \times 10^{6} \times (10 \times 365 \times 24 \times 3600)$
$E = 10^{9} \times (10 \times 365 \times 24 \times 3600)$
$E = 3.15 \times 10^{17} J$
When efficiency is $10 %$ the energy will be-
$η = \frac{Energy output}{Energy input}$
$E_{Input} = \frac{3.15 \times 10^{17}}{\frac{10}{100}} = 3.15 \times 10^{18} J$
$∵ E_{fission} = 200 \times 10^{6} \times 1.6 \times 10^{- 19}$
$E_{fission} = 3.2 \times 10^{- 11} J$
We know that,
$Total number of fission = \frac{E_{In}}{E_{fission}}$
$= \frac{3.15 \times 10^{18}}{3.2 \times 10^{- 11}}$
$= 9.8 \times 10^{28}$
We know that,
$Moles (n) = \frac{Total number of fission}{N_{A}}$
$n = \frac{9.8 \times 10^{28}}{6.02 \times 10^{26}}$
$n = 163.7$
Mass of uranium required is given by-
$m = 163.7 \times 235$
$m = 38.2 \times 10^{3} kg$

Manipal UGET-2015

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation: