Answered

Discover answers to your questions with Westonci.ca, the leading Q&A platform that connects you with knowledgeable experts. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.

(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].
We hope this was helpful. Please come back whenever you need more information or answers to your queries. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.