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Karnataka Board PUCPUC Science 2nd PUC Class 12

Find the Q-value and the kinetic energy of the emitted α-particle in the α-decay of Rn86220Rn.

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Question

Find the Q-value and the kinetic energy of the emitted α-particle in the α-decay of `""_86^220"Rn"`.

Given `"m"(""_88^226"Ra")` = 226.02540 u, `"m"(""_86^222 "Rn")` = 222.01750 u, 

`"m"(""_86^220 "Rn")`= 220.01137 u, `"m"(""_84^216 "Po")`= 216.00189 u.

Numerical
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Solution

Alpha particle decay of `(""_86^220"Rn")` is shown by the following nuclear reaction.

\[\ce{^220_86Rn -> ^216_84Po + ^4_2He}\]

It is given that:

Mass of `(""_86^220 "Rn")` = 220.01137 u

Mass of `(""_84^216 "Po")` = 216.00189 u

∴ Q-value = [220.01137 - (216.00189 +.00260)] × 931.5

≈ 641 MeV

Kinetic energy of the α-particle = `((220-4)/220) xx 6.41`

= 6.29 MeV

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Chapter 13: Nuclei - Exercise [Page 463]

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NCERT Physics Part I and II [English] Class 12
Chapter 13 Nuclei
Exercise | Q 12.2 | Page 463
NCERT Physics Part I and II [English] Class 12
Chapter 13 Nuclei
Exercise | Q 13.12 (b) | Page 463

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