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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.

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#### 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) c^{2}

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 c^{2}

= 0.005297 u c^{2}

But 1 u = 931.5 MeV/c^{2}

∴ 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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