SATHEE: Work Energy And Power Question 88

Work Energy And Power Question 88

Question: A $^{238} U$ nucleus decays by emitting an alpha particle of speed $v m s^{- 1}$ . The recoil speed of the residual nucleus is (in $m s^{- 1}$ ) [CBSE PMT 1995; AIEEE 2003]

Options:

A) $- 4 v / 234$

B) $v / 4$

C) $- 4 v / 238$

D) $4 v / 238$

Show Answer

Answer:

Correct Answer: A

Solution:

Initially 238U nucleus was at rest and after decay its part moves in opposite direction. According to conservation of momentum $4 v + 234 V$ = 238 × 0 therefore $V = - \frac{4 v}{234}$

Work Energy And Power Question 88

Question: A $^{238} U$ nucleus decays by emitting an alpha particle of speed $v m s^{- 1}$ . The recoil speed of the residual nucleus is (in $m s^{- 1}$ ) [CBSE PMT 1995; AIEEE 2003]

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

Work Energy And Power Question 88

Question: A 238U nucleus decays by emitting an alpha particle of speed vms−1 . The recoil speed of the residual nucleus is (in ms−1 ) [CBSE PMT 1995; AIEEE 2003]

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: A $^{238} U$ nucleus decays by emitting an alpha particle of speed $v m s^{- 1}$ . The recoil speed of the residual nucleus is (in $m s^{- 1}$ ) [CBSE PMT 1995; AIEEE 2003]