Question

(a) Calculate the energy released by the alpha decay of 222 86Rn.

(b) Calculate the energy of the alpha particle.

(c) What is the energy of the recoil polonium atom

Answer 1

Answer:

a). The energy released by the alpha decay of 222 Rn is to 218 Po :

= 5.596 MeV

b).The energy of the alpha particle is 5.5 MeV

c) The recoil energy of Po = 0.096 MeV

Explanation:

The equation for the alpha decay of Rn to Polonium is:

$_{86}^{222}\textrm{Rn}\rightarrow$ $_{2}^{4}\textrm{He}+_{84}^{218}\textrm{Po}$

The energy of the decay process can be calculated using:

$\Delta E=\Delta mc^{2}$

$\Delta m$ = Change in the mass

Mass of Po = 218.008965 u

Mass of He = 4.002603 u

Mass of Rn = 222.017576 u

$\Delta m$ = mass of Po + mass of He - mass of Rn

=  4.002603 + 218.008965-222.017576  

= - 0.006008

$\Delta E=m(931.5MeV)$

$\Delta E=-0.006008\times 931.5MeV$

= -5.596 MeV

The negative sign means energy is released during the process.

b) The energy of the alpha particle is :

$\frac{Po\ mass }{Rn\ mass}\times E$

$\frac{218.0089}{222.0175}\times \Delta E$

$\frac{218.0089}{222.0175}\times 5.596$

= 5.494 MeV

= 5.5 MeV

c).

Recoil energy: When the parent nucleus is at rest before the decay then , there must be the some recoil of daughter nucleus to conserve the momentum.This is termed as recoil energy.

The energy of the recoil polonium atom :

The formula for recoil energy is :

The total energy - the kinetic energy

= 5.596 - 5.5

= 0.096 MeV