Ncert Exercises & Solutions

13.12 Find the Q-value and the kinetic energy of the emitted α-particle in the α-decay of (a) \({ }_{88}^{226} \mathrm{Ra}\) and (b) \({ }_{86}^{220} \mathrm{Rn}\).

Given: m (\({ }_{88}^{226} \mathrm{Ra}\)) = 226.02540 u, m (\({ }_{86}^{222} \mathrm{Rn}\)) = 222.01750 u,

m (\({ }_{86}^{220} \mathrm{Rn}\)) = 220.01137 u, m (\({ }_{84}^{216} \mathrm{Po}\)) = 216.00189 u.

Hint: Q-value is given by mass defect as, \(E=\Delta mc^{2}\).
(a)
Step 1: Find the reaction for α-decay of \({ }_{88}^{226} \mathrm{Ra}\).
Alpha particle decay of

{}_{88}^{226}R a

emits a helium nucleus. As a result, its atomic mass number reduces to (226-4) 222 and its atomic number reduces to (88-2) 86. This is shown in the following nuclear reaction:

{}_{86}^{226}R a \to {}_{86}^{222}R a + {}_{2}^{4}H e

Step 2: Find the Q-value and kinetic energy of α-particle in α-decay of \({ }_{88}^{226} \mathrm{Ra}\).
Q-value of emitted

α

-particle=(Sum of initial mass - Sum of final mass)x

c^{2}

where, c=speed of light
It is given that:

m ({}_{86}^{226}R a) = 226.02540 u

m ({}_{86}^{222}R n) = 222.01750 u

m ({}_{2}^{4}H e) = 4.002603 u

Q - v a l u e = [226.02540 - (222.01750 + 4.002603)] u c^{2} = 0.005297 u c^{2}

B u t 1 u = 931.5 M e V / c^{2}

∴ Q = 0.005297 \times 931.5 \approx 4.94 M ev

K i n e t i c e n e r g y o f t h e α - p a r t i c l e = \frac{M a s s n u m b e r a f t e r d e c a y}{M a s s n u m b e r b e f o r e d e c a y} \times Q = \frac{222}{226} \times 4.94 = 4.85 M eV

(b)
Step 3: Find the reaction for α-decay of \({ }_{86}^{220} \mathrm{Rn}\).
Alpha particle decay of \({ }_{86}^{220} \mathrm{Rn}\)

{}_{86}^{220}{Rn} \to {}_{84}^{216}P o + {}_{e}^{4}H e

Step 4: Find the Q-value and kinetic energy of α-particle in α-decay of \({ }_{86}^{220} \mathrm{Rn}\).
It is given that:

Mass of

{}_{86}^{220}R n

=220.01137 u and mass of

{}_{84}^{216}P o

=216.00189 u

∴ Q - v a l u e = [220.01137 - (216.0019 + 4.00260)] \times 931.5 \approx 641 M e V

The kinetic energy of the

α

-particle =

(\frac{220 - 4}{220}) \times 6.41 = 6.29 M e V

NEET Information

courses

Company

Botany Questions

Chemistry Questions

Physics Questions

Zoology Questions