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 - u-A- 5-58 MeVB- 4-72 MeVC- 5-03 MeVD- 3-34 MeV

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Byju's Answer

Standard IX

Physics

Alpha Decay

So, in the ra...

Question

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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Similar questions

Q. Don't hate me for asking this question, as it will need you to work your calculator! So, in the radioactive decay -

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Q. The kinetic energy of the

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_{94}^{238} P u \to_{92}^{234} U +

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Q. Radiometric dating is a technique used to find the age of materials on Earth, using the concept of radioactive decay (by measuring the relative abundance of the parent and daughter nuclei). One example of this is Uranium-Thorium dating, used for dating sediments, wherein Uranium-238 goes through a single alpha decay to give Thorium-234:

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^{238} U \to^{234} T h + α

What will the Q-value of the process be?
Things you will need:
1. Atomic masses:

m (^{238} U) = 238.0508 u; m (^{234} U) = 234.04363 u; m (^{4} H e) = 4.00260 u

.
2. Constant:

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So, in the radioactive decay - 238Pu→234U+α, What will be the kinetic energy of the outgoing α-particle? Things you will need:

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: