Question

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

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