(A) Power $P = 1 \, kW = 1000 \, W = 1000 \, J/s$.Energy released per fission $E = 200 \, MeV = 200 \times 10^6 \times 1.6 \times 10^{-19} \, J = 3.2

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(A) Power $P = 1 \, kW = 1000 \, W = 1000 \, J/s$.Energy released per fission $E = 200 \, MeV = 200 \times 10^6 \times 1.6 \times 10^{-19} \, J = 3.2 \times 10^{-11} \, J$.Let $n$ be the number of fissions per second.The total power produced is $P = n \times E$.Therefore, $n = P / E$.$n = 1000 / (3.2 \times 10^{-11}) = 0.3125 \times 10^{14} = 3.125 \times 10^{13} \, \text{fissions per second}$.

Class 12 Physics Nuclei Nuclear Fission, Fusion and Nuclear Reactor

Medium

If $200 \, MeV$ energy is released in the fission of a single ${U^{235}}$ nucleus, the number of fissions required per second to produce $1 \, kW$ power shall be (Given $1 \, eV = 1.6 \times 10^{-19} \, J$)

A
$3.125 \times 10^{13}$
B
$3.125 \times 10^{14}$
C
$3.125 \times 10^{15}$
D
$3.125 \times 10^{16}$

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Physics Questions

Chapter

Nuclei

Subtopic

Nuclear Fission, Fusion and Nuclear Reactor

Light energy emitted by stars is due to

Easy

View Solution

When a uranium isotope ${ }_{92}^{235} U$ is bombarded with a neutron,it generates ${ }_{36}^{89} Kr$,three neutrons and:

EasyNEET 2020

View Solution

$A$ nucleus with mass number $240$ breaks into two fragments each of mass number $120$. The binding energy per nucleon of the unfragmented nucleus is $7.6 \, MeV$,while that of the fragments is $8.5 \, MeV$. The total gain in the binding energy in the process is (in $MeV$):

MediumNEET 2021

View Solution

Let $E_1$ and $E_2$ be the binding energies of two nuclei $A$ and $B.$ It is observed that two nuclei of $A$ combine together to form a $B$ nucleus. This observation is correct only if

Medium

View Solution

The energy released by the fission of one uranium atom is $200 MeV$. The number of fissions per second required to produce $3.2 W$ of power is (Take $1 eV = 1.6 \times 10^{-19} J$).

EasyWBJEE 2010

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

If $200 \, MeV$ energy is released in the fission of a single ${U^{235}}$ nucleus, the number of fissions required per second to produce $1 \, kW$ power shall be (Given $1 \, eV = 1.6 \times 10^{-19} \, J$)

Similar Questions

Light energy emitted by stars is due to

When a uranium isotope ${ }_{92}^{235} U$ is bombarded with a neutron,it generates ${ }_{36}^{89} Kr$,three neutrons and:

$A$ nucleus with mass number $240$ breaks into two fragments each of mass number $120$. The binding energy per nucleon of the unfragmented nucleus is $7.6 \, MeV$,while that of the fragments is $8.5 \, MeV$. The total gain in the binding energy in the process is (in $MeV$):

Let $E_1$ and $E_2$ be the binding energies of two nuclei $A$ and $B.$ It is observed that two nuclei of $A$ combine together to form a $B$ nucleus. This observation is correct only if

The energy released by the fission of one uranium atom is $200 MeV$. The number of fissions per second required to produce $3.2 W$ of power is (Take $1 eV = 1.6 \times 10^{-19} J$).

Vedclass Test Series

Exam Paper Generator

Online Exam Module