The average energy released per fission for t | vedclass

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

Question

(B) The molar mass of $_{92}^{235} U$ is $235 	ext{ g/mol}$.Number of moles in $47 	ext{ g}$ of $_{92}^{235} U$ is $n = \frac{47 	ext{ g}}{235 	ex

Vedclass · Accepted Answer

$228$

Vedclass · Answer

(B) The molar mass of $_{92}^{235} U$ is $235 	ext{ g/mol}$.Number of moles in $47 	ext{ g}$ of $_{92}^{235} U$ is $n = \frac{47 	ext{ g}}{235 	ext{ g/mol}} = 0.2 	ext{ moles} = \frac{1}{5} 	ext{ moles}$.Total number of atoms $N = n 	imes N_A = \frac{1}{5} 	imes 6 	imes 10^{23} = 1.2 	imes 10^{23} 	ext{ atoms}$.Energy released per fission is $190 	ext{ MeV}$.Total energy released $= N 	imes 190 	ext{ MeV} = (1.2 	imes 10^{23}) 	imes 190 	ext{ MeV} = 228 	imes 10^{23} 	ext{ MeV}$.Comparing this with $\alpha 	imes 10^{23} 	ext{ MeV}$, we get $\alpha = 228$.

Similar Questions

The energy released when one nucleus of ${ }_{92} U^{235}$ undergoes fission is $188 MeV$. The energy released when $100 g$ of ${ }_{92} U^{235}$ undergoes fission is:

Two lithium nuclei in a lithium vapour at room temperature do not combine to form a carbon nucleus because

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 $O = 15.9994 \, a.m.u.$]

The source of the Sun's energy is

Which of the following isotopes is used for the treatment of cancer?

Vedclass Test Series

Exam Paper Generator

Online Exam Module