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 =

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

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