The energy released per fission of a nucleus | vedclass

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

Question

(C) The number of moles of ${}^{240}X$ is given by $n = \frac{	ext{mass}}{	ext{molar mass}} = \frac{120 \ g}{240 \ g/mol} = 0.5 \ mol$.The number of

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(C) The number of moles of ${}^{240}X$ is given by $n = \frac{	ext{mass}}{	ext{molar mass}} = \frac{120 \ g}{240 \ g/mol} = 0.5 \ mol$.The number of atoms (nuclei) in $0.5 \ mol$ is $N = n 	imes N_A = 0.5 	imes 6 	imes 10^{23} = 3 	imes 10^{23} \ 	ext{atoms}$.The energy released per fission is $E_{fission} = 200 \ MeV$.The total energy released is $E_{total} = N 	imes E_{fission} = (3 	imes 10^{23}) 	imes (200 \ MeV) = 600 	imes 10^{23} \ MeV = 6 	imes 10^{25} \ MeV$.Thus,the value is $6$.

The energy released per fission of a nucleus of ${}^{240}X$ is $200 \ MeV$. The energy released if all the atoms in $120 \ g$ of pure ${}^{240}X$ undergo fission is $........ \times 10^{25} \ MeV$. (Given $N_A = 6 \times 10^{23}$)

Similar Questions

$A$ nuclear reactor delivers a power of $10^9 \,W$. The amount of fuel consumed by the reactor in one hour is: (in $\,g$)

What is the amount of energy released by deuterium and tritium fusion (in $,MeV$)?

In a nuclear fission reaction of an isotope of mass $M$,three similar daughter nuclei of the same mass are formed. The speed of a daughter nucleus in terms of mass defect $\Delta M$ will be:

The fission of $_{92}^{235}U$ can be triggered by the absorption of a slow neutron by a nucleus. Similarly,a slow proton can also be used. This statement is

The energy released by the fission of one uranium atom is $200 \text{ MeV}$. The number of fissions per second required to produce $6.4 \text{ W}$ power is $\qquad$ .

Vedclass Test Series

Exam Paper Generator

Online Exam Module