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(b) The radioactive isotope [tex][tex]${ }_{84}^{210} Po$[/tex][/tex] emits alpha particles of single energy, the product nuclei being radioactive isotope [tex][tex]${ }_{82}^{206} Po$[/tex][/tex].

(i) Using the data below, calculate the energy in [tex][tex]$MeV$[/tex][/tex] released in each disintegration.

(ii) Calculate the kinetic energy of the alpha particle.

(Masses of [tex][tex]${ }_{84}^{210} Po =209.9367 u$[/tex][/tex]; [tex]{ }_{82}^{206} Po =205.9294 u ; \quad{ }_2^4 \alpha=3.894 u[/tex])

Sagot :

GinnyAnswer
Certainly! Let's solve this problem step-by-step.

### Given Data:
- Mass of [tex]\( {}_{84}^{210}Po = 209.9367 \)[/tex] atomic mass units (u)
- Mass of [tex]\( {}_{82}^{206}Po = 205.9294 \)[/tex] atomic mass units (u)
- Mass of [tex]\( {}_{2}^{4}\alpha = 3.894 \)[/tex] atomic mass units (u)

### (i) Energy Released in Each Disintegration:

First, we need to calculate the mass defect of this reaction. The mass defect ([tex]\( \Delta m \)[/tex]) is the difference between the mass of the parent nucleus and the total mass of the daughter nucleus and the emitted alpha particle.

[tex]\[ \Delta m = \text{mass of } {}_{84}^{210}Po - (\text{mass of } {}_{82}^{206}Po + \text{mass of } {}_2^4\alpha) \][/tex]

Substitute the given values:

[tex]\[ \Delta m = 209.9367 \, \text{u} - (205.9294 \, \text{u} + 3.894 \, \text{u}) \][/tex]
[tex]\[ \Delta m = 209.9367 \, \text{u} - 209.8234 \, \text{u} \][/tex]
[tex]\[ \Delta m = 0.1133 \, \text{u} \][/tex]

Next, we need to convert this mass defect into energy. We use Einstein's equation [tex]\( E = mc^2 \)[/tex], and the conversion factor [tex]\( 1 \, \text{u} = 931.5 \, \text{MeV}/c^2 \)[/tex].

[tex]\[ E = \Delta m \times 931.5 \, \text{MeV}/c^2 \][/tex]

Substitute [tex]\(\Delta m\)[/tex]:

[tex]\[ E = 0.1133 \, \text{u} \times 931.5 \, \text{MeV}/\text{u} \][/tex]
[tex]\[ E = 105.539 \, \text{MeV} \][/tex]

So, the energy released in each disintegration is approximately [tex]\( 105.539 \, \text{MeV} \)[/tex].

### (ii) Kinetic Energy of the Alpha Particle:

Assuming that the daughter nucleus [tex]\( {}_{82}^{206}Po \)[/tex] is at rest after the emission, the kinetic energy of the alpha particle will be equal to the total energy released in the disintegration.

Therefore, the kinetic energy of the alpha particle is:

[tex]\[ \text{Kinetic Energy of } \alpha \text{ particle} = 105.539 \, \text{MeV} \][/tex]

### Summary:

(i) The energy released in each disintegration is approximately [tex]\( 105.539 \, \text{MeV} \)[/tex].
(ii) The kinetic energy of the alpha particle is approximately [tex]\( 105.539 \, \text{MeV} \)[/tex].
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