SATHEE: Atoms And Nuclei Question 212

Atoms And Nuclei Question 212

Question: 200 MeV of energy may be obtained per fission of $U^{235}$ . A reactor is generating 1000 kW of power. The rate of nuclear fission in the reactor is [MP PET 1995]

Options:

A) 1000

B) $2 \times 10^{8}$

C) $3.125 \times 10^{16}$

D) 931

Show Answer

Answer:

Correct Answer: C

Solution:

Power
$\Rightarrow {(\frac{1}{2})}^{4} = {(\frac{1}{2})}^{t / 48} \Rightarrow t = 192 h o u r .$ Rate of nuclear fission $= \frac{10^{6}}{200 \times 1.6 \times 10^{- 13}}$ = 3.125 ´ 1016.

Atoms And Nuclei Question 212

Question: 200 MeV of energy may be obtained per fission of $U^{235}$ . A reactor is generating 1000 kW of power. The rate of nuclear fission in the reactor is [MP PET 1995]

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Atoms And Nuclei Question 212

Question: 200 MeV of energy may be obtained per fission of U235 . A reactor is generating 1000 kW of power. The rate of nuclear fission in the reactor is [MP PET 1995]

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu

Question: 200 MeV of energy may be obtained per fission of $U^{235}$ . A reactor is generating 1000 kW of power. The rate of nuclear fission in the reactor is [MP PET 1995]