SATHEE: Atoms And Nuclei Question 236

Atoms And Nuclei Question 236

Question: To generate a power of 3.2 mega watt, the number of fissions of $U^{235}$ per minute is (Energy released per fission = 200MeV, $1 e V = 1.6 \times 10^{- 19} J)$ [EAMCET (Engg.) 2000]

Options:

A) $6 \times 10^{18}$

B) $6 \times 10^{17}$

C) $10^{17}$

D) $6 \times 10^{16}$

Show Answer

Answer:

Correct Answer: A

Solution:

Number of fissions per second $= \frac{P o w e r o u t p u t}{E n e r g y r e l e a s e d p e r f i s s i o n}$ $= \frac{3.2 \times 10^{6}}{200 \times 10^{6} \times 1.6 \times 10^{- 19}} = 1 \times 10^{17}$
$\Rightarrow$ Number of fission per minute $= 60 \times 10^{17} = 6 \times 10^{18}$

Atoms And Nuclei Question 236

Question: To generate a power of 3.2 mega watt, the number of fissions of $U^{235}$ per minute is (Energy released per fission = 200MeV, $1 e V = 1.6 \times 10^{- 19} J)$ [EAMCET (Engg.) 2000]

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 236

Question: To generate a power of 3.2 mega watt, the number of fissions of U235 per minute is (Energy released per fission = 200MeV, 1eV=1.6×10−19J) [EAMCET (Engg.) 2000]

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

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Question: To generate a power of 3.2 mega watt, the number of fissions of $U^{235}$ per minute is (Energy released per fission = 200MeV, $1 e V = 1.6 \times 10^{- 19} J)$ [EAMCET (Engg.) 2000]