Two radioactive substances A and B of mass numbers 200 and 212 respectively, shows spontaneous $$\alpha$$-decay with same Q value of 1 MeV. The ratio of energies of $$\alpha$$-rays produced by A and B is __________.

Solution :

In $$\alpha$$-decay :

$$Q = K_\alpha + K_D$$

where :

$$K_\alpha$$ = kinetic energy of $$\alpha$$-particle

$$K_D$$ = kinetic energy of daughter nucleus

Using conservation of momentum :

$$p_\alpha = p_D$$

Therefore,

$$\frac{K_\alpha}{K_D} = \frac{M_D}{M_\alpha}$$

Hence,

$$K_\alpha = Q\left(\frac{M_D}{M_D+M_\alpha}\right)$$

For substance $$A$$ :

Mass number of parent nucleus :

$$200$$

Daughter nucleus mass number :

$$196$$

Therefore,

$$K_A = Q\left(\frac{196}{196+4}\right)$$

$$= Q\left(\frac{196}{200}\right)$$

$$= \frac{49Q}{50}$$

For substance $$B$$ :

Mass number of parent nucleus :

$$212$$

Daughter nucleus mass number :

$$208$$

Therefore,

$$K_B = Q\left(\frac{208}{208+4}\right)$$

$$= Q\left(\frac{208}{212}\right)$$

$$= \frac{52Q}{53}$$

Required ratio :

$$\frac{K_A}{K_B} = \frac{49/50}{52/53}$$

$$= \frac{49 \times 53}{50 \times 52}$$

$$= \frac{2597}{2600}$$

Final Answer :

$$\frac{2597}{2600}$$