Find the Q-value and the Kinetic Energy of the Emitted α-particle in the α-decay of _88^226 Ra and _86^220rn - Physics


Find the Q-value and the kinetic energy of the emitted α-particle in the α-decay of `""_88^226 "Ra"`.

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.



Alpha particle decay of `""_88^226"Ra"` emits a helium nucleus. As a result, its mass number reduces to (226 − 4) 222 and its atomic number reduces to (88 − 2) 86. This is shown in the following nuclear reaction.

\[\ce{^226_88 Ra -> ^222_86 Ra + ^4_2He}\]

Q-value of

emitted α-particle = (Sum of initial mass − Sum of final mass) c2


c = Speed of light

It is given that:

`"m"(""_88^226"Ra")` = 226.02540 u

`"m"(""_86^222"Rn")` = 222.01750 u

`"m"(""_2^4"He")` = 4.002603 u

Q-value = [226.02540 − (222.01750 + 4.002603)] u c2 
= 0.005297 u c2

But 1 u = 931.5 MeV/c2

∴ Q = 0.005297 × 931.5 ≈ 4.94 MeV

Kinetic energy of the α-particle  = `("Mass number after decay"/"Mass number before decay") xx "Q"`

` = 222/226 xx 4.94 = 4.85 " MeV"`

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Chapter 13: Nuclei - Exercise [Page 463]


NCERT Physics Class 12
Chapter 13 Nuclei
Exercise | Q 13.12 (a) | Page 463
NCERT Physics Class 12
Chapter 13 Nuclei
Exercise | Q 12.1 | Page 463


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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, 

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