113. Energy released in the fission of a sing | vedclass

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

Question

(A) Power $P = 5 	ext{ mW} = 5 	imes 10^{-3} 	ext{ W}$.Energy released per fission $E = 200 	ext{ MeV} = 200 	imes 10^6 	imes 1.6 	imes 10^{-19

Vedclass · Accepted Answer

$1.56 	imes 10^8$

Vedclass · Answer

(A) Power $P = 5 	ext{ mW} = 5 	imes 10^{-3} 	ext{ W}$.Energy released per fission $E = 200 	ext{ MeV} = 200 	imes 10^6 	imes 1.6 	imes 10^{-19} 	ext{ J} = 3.2 	imes 10^{-11} 	ext{ J}$.Let $n$ be the number of fissions per second.Power $P = n 	imes E$.Therefore, $n = \frac{P}{E} = \frac{5 	imes 10^{-3} 	ext{ J/s}}{3.2 	imes 10^{-11} 	ext{ J/fission}}$.$n = \frac{5}{3.2} 	imes 10^8 = 1.5625 	imes 10^8 	ext{ fissions/s}$.Thus, the number of fissions per second is $1.56 	imes 10^8$.

$113$. Energy released in the fission of a single uranium nucleus is $200 \text{ MeV}$. Then the number of fissions per second to produce $5 \text{ mW}$ power is

Similar Questions

The atomic power station at Tarapore has a generating capacity of $200\, MW$. The energy generated in a day by this station is:

The large-scale destruction that would be caused due to the use of nuclear weapons is called:

The following is not used as a nuclear fuel.

Nuclear fission is possible because the binding energy per nucleon ..........

If all the helium in a star is converted into oxygen in its core,the energy released per oxygen nucleus is...........$MeV$?
[Mass of $He = 4.0026 \, a.m.u.$,Mass of $O = 15.9994 \, a.m.u.$]

Vedclass Test Series

Exam Paper Generator

Online Exam Module