Question 74: Atoms & Nuclei

Question

The mass of a nucleus $$_Z^AX$$ is less than the sum of the masses of (A-Z) number of neutrons and Z number of protons in the nucleus. The energy equivalent to the corresponding mass difference is known as the binding energy of the nucleus. A heavy nucleus of mass M can break into two light nuclei of masses m1 and m2 only if (m1 + m2)  M'. The masses of some neutral atoms are given in the table below :

$$_1^1H$$
1.007825 u
$$_1^2H$$
2.014102 u

$$_3^6Li$$
6.015123 u
$$_3^7Li$$
7.016004 u

$$_{64}^{152}Gd$$
151.919803 u
$$_{82}^{206}Pb$$
205.974455 u

$$_1^3H$$
3.016050 u
$$_2^4He$$
4.002603 u

$$_{30}^{70}Zn$$
69.925325 u
$$_{34}^{82}Se$$
81.916709 u

$$_{83}^{209}Bi$$
208.980388 u
$$_{84}^{210}Po$$
209.982876 u

(1 u = 932 MeV/c2)

Accepted Answer

The alpha decay of $$_{84}^{210}Po$$ is given by
$$_{84}^{210}Po 	o _{82}^{206}Pb + _2^4He$$
The energy released during this process is
$$Q = ({M_{Po}} - {M_{Pb}} - {M_{He}}){c^2}$$
$$ = (209.982876 - 205.974455 - 4.002603)\,u 	imes {c^2}$$
$$ = (0.005818\,u){c^2} = (0.005818\,u) 	imes 932\,MeV$$
$$ = 5.422\,MeV = 5422\,keV$$
Kinetic energy of a particle, $${K_\alpha } = {{(A - 4)Q} \over A}$$
$${K_\alpha } = {{(210 - 4)} \over {210}} 	imes 5422\,keV = {{206} \over {210}} 	imes 5422\,keV = 5319\,keV$$

Answer

5319

Answer

5422

Answer

5707

Answer

5818

Atoms & Nuclei

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