Test Index

NCERT Class XII Chapter
Nuclei
Questions With Solutions

© examsnet.com
Question : 12 of 31
Marks: +1, -0
Find the Q-value and the kinetic energy of the emitted α-particle in the α-decay of (a)  88226Ra\,{}^{226}_{88}\mathrm{Ra} and (b)  86220Rn\,{}^{220}_{86}\mathrm{Rn}
Given m (88226Ra)({}^{226}_{88}\mathrm{Ra}) = 226.02540 u , m ( 86222Rn)({\,{}^{222}_{86}\mathrm{Rn}}) = 222.01750 u ,
m ( 86220Rn)(\,{}^{220}_{86}\mathrm{Rn}) = 220.01137u, m (84216Po)(^{216}_{84}\mathrm{Po}) = 216.00189u, and
mαm_{\alpha} = 4.00260 u
Solution:  
(a) α-decay of  88226Ra\,{}^{226}_{88}\mathrm{Ra}
 88226Ra\,{}^{226}_{88}\mathrm{Ra} →  86222Rn+ 24He+Q\,{}^{222}_{86}\mathrm{Rn}+\,{}^{4}_{2}\mathrm{He}+Q
so, Q value
Q =
[m( 88226Ra)−m( 86222Rn)−m( 24He)]c2\left[ m\left(\,{}^{226}_{88}\mathrm{Ra}\right)-m\left(\,{}^{222}_{86}\mathrm{Rn}\right)-m\left(\,{}^{4}_{2}\mathrm{He}\right) \right]c^2
Q = [226.02540 – 222.01750 – 4.00260] × 931.5 MeV
Q = 0.0053 × 931.5 MeV = 4.937 MeV
Kinetic energy of emitted α-particle
KαK_{\alpha} = QA\frac{Q}{A} (A - 4) or KαK_{\alpha} = 4.937226\frac{4.937}{226} × (226 - 4) MeV
KαK_{\alpha} = 4.85 MeV
(b) α-decay of  86220Rn\,{}^{220}_{86}\mathrm{Rn}
 86220Rn\,{}^{220}_{86}\mathrm{Rn} →  84216Po+ 24He+Q\,{}^{216}_{84}\mathrm{Po}+\,{}^{4}_{2}\mathrm{He}+Q
Q =
[m( 86220Rn)−m( 84216Po)−m( 24He)]c2\left[ m\left(\,{}^{220}_{86}\mathrm{Rn}\right)-m\left(\,{}^{216}_{84}\mathrm{Po}\right)-m\left(\,{}^{4}_{2}\mathrm{He}\right) \right]c^2
Q = [220.01137 – 216.00189 – 4.00260] 931.5 MeV = 6.41 MeV
Kinetic energy of emitted α particle
KαK_{\alpha} = QA\frac{Q}{A} (A - 4) = 6.41220\frac{6.41}{220} × (220 - 4) = 6.29 MeV
© examsnet.com
Go to Question:
Prev Question
Next Question