# 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

Numerical

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.

#### Solution

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

Where,

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"

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

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