Two radioactive substances A and B of mass nu | vedclass

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(C) The kinetic energy of the $\alpha$-particle is given by $K_{\alpha} = \frac{A-4}{A} Q$,where $A$ is the mass number of the parent nucleus.For subs

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$2597$/$2600$

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(C) The kinetic energy of the $\alpha$-particle is given by $K_{\alpha} = \frac{A-4}{A} Q$,where $A$ is the mass number of the parent nucleus.For substance $A$ with mass number $A_1 = 200$,the kinetic energy of the $\alpha$-particle is $K_A = \frac{200-4}{200} Q = \frac{196}{200} Q$.For substance $B$ with mass number $A_2 = 212$,the kinetic energy of the $\alpha$-particle is $K_B = \frac{212-4}{212} Q = \frac{208}{212} Q$.The ratio of the energies is $\frac{K_A}{K_B} = \frac{196}{200} 	imes \frac{212}{208}$.Simplifying the expression: $\frac{196 	imes 212}{200 	imes 208} = \frac{41552}{41600}$.Dividing both numerator and denominator by $16$,we get $\frac{2597}{2600}$.

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

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