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प्रश्न
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.
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उत्तर
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
संबंधित प्रश्न
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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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