Question

Obtain the maximum kinetic energy of $\beta$ -particles, and the radiation frequencies of $\gamma$ decays in the decay scheme shown.

You are given that
$m{(}^{198}Au)=197.968233u$
$m{(}^{198}Hg)=197.966760u$

Answer 1

It can be observed from the given $\gamma$ decay diagram that ${\gamma }_1$ decays from the 1.088 MeV energy level to the 0 MeV energy level.
Hence, the energy corresponding to ${\gamma }_1$ decays is given as

$E_1=1.088-0=1.088MeV$

$h{v}_1=1.088\times 1.6\times {10}^{-19}\times {10}^{6}J$

Where,

h= Plank's constant and $v_1$ = frequency of radiation radiated by ${\gamma }_1$ decay.

$\therefore v_1=\frac{E_1}{h}=\frac{1.088\times 1.6\times {10}^{19}\times {10}^6}{6.6\times {10}^{-34}}=2.637\times {10}^{20}Hz$

It can be observed the given gamma decay diagram that ${\gamma }_2$ decays from the 0.412 MeV energy level to the 0 MeV energy level.

Hence, the energy corresponding to ${\gamma }_2$ decay is given as

$E_2=0.412-0=0.412MeV$

$h{v}_2=0.412\times 1.6\times {10}^{-19}\times {10}^{6}J$

Where $v_2$ is the frequency of the radiation radiated by ${\gamma }_2$ decay.

$\therefore v_2=\frac{E_2}{h}=9.988\times {10}^{19}Hz$

It can be observed from the given gamma decay diagram that ${\gamma }_3$ decays from the 1.088 MeV energy level to the 0.412 MeV energy level. Hence, the energy corresponding to ${\gamma }_3$ is given as

$E_3=1.088-0.412=m0.676MeVJ$

$h{v}_3=0.676\times {10}^{-19}\times {10}^{6}J$

Where $v_3=$ frequency of the radiation radiated by ${\gamma }_3$ decay

$\therefore v_3=\frac{E_3}{h}=1.639\times {10}^{20}Hz$

Now mass of $Au$ is 197.968 u and mass of $Hg$ = 197.9667 u

$1u=931.5MeV/c^2$

Energy of the highest level is given as :

$E=\left[m{(}_{78}^{198}Au\right)-m{(}_{80}^{190}Hg\left)\right]=197.968233-197.96676=0.001473u=0.001473\times 931.5=1.3720995MeV$

${\beta }_1$ decays from the 1.3720995 MeV level to the 1.088 MeV level

$\therefore$ maximum kinetic energy of the ${\beta }_1$ particle $=1.3720995-1.088=0.2840995MeV$

${\beta }_2$ decays from the 1.3720995 MeV level to the 0.412 MeV level

$\therefore$ Maximum kinetic energy of the ${\beta }_2$ particle = $1.3720995-0.412=0.9600995$