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

Numerical

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.

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

Concept: Atomic Masses and Composition of Nucleus
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APPEARS IN

NCERT Physics Part 1 and 2 Class 12
Chapter 13 Nuclei
Exercise | Q 13.12 (b) | Page 463
NCERT Class 12 Physics Textbook
Chapter 13 Nuclei
Exercise | Q 12.2 | Page 463
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