Don-t hate me for asking this question- as it will need you to work your calculator- So- in the radioactive decay 238 P u -234 U -What will be the kinetic energy of the outgoing -particle- Things you will need- Atomic masses- m 238 P u -238-04955 u - m 234 U -234-04095 u - m 4 He -4-002603 u- Constant- C 2-931 MeV - uA- 5-58 MeVB- 4-72 MeVC- 5-03 MeVD- 3-34 MeV

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Byju's Answer

Standard XII

Physics

Law of Radioactivity

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Question

Don't hate me for asking this question, as it will need you to work your calculator! So, in the radioactive decay -
$^{238} P u \to^{234} U + α,$
What will be the kinetic energy of the outgoing α-particle?
Things you will need:

5.58 MeV

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4.72 MeV

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5.03 MeV

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3.34 MeV

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Solution

The correct option is A 5.58 MeV
This apparent loss in mass of the system of decay products will manifest itself as kinetic energy of the alpha particle. This energy is called the Q value of the decay. Therefore,
$K i n e t i c e n e r g y o f α = Q v a l u e = [m (^{238} P u) - {m (^{234} U) + m (^{4} H e)}] \times c^{2}$
$= [238.04955 u - (234.04095 u + 4.002603 u)] \times 931 \frac{M e V}{u}$
$= \begin{matrix} 5.58 M e V \end{matrix} .$

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Don't hate me for asking this question, as it will need you to work your calculator! So, in the radioactive decay - 238Pu→234U+α, What will be the kinetic energy of the outgoing α-particle? Things you will need:

Don't hate me for asking this question, as it will need you to work your calculator! So, in the radioactive decay -
$^{238} P u \to^{234} U + α,$
What will be the kinetic energy of the outgoing α-particle?
Things you will need: