Under certain circumstances, a nucleus can decay by emitting a particle more massive than an α-particle.

Solution:

Take a ${ }_{6}^{14} \mathrm{C}$ emission nuclear reaction:

${ }_{88}^{223} \mathrm{Ra} \longrightarrow{ }_{82}^{209} \mathrm{~Pb}+{ }_{6}^{14} \mathrm{C}$

We know that:

Mass of ${ }_{88}^{223} \mathrm{Ra}, m_{1}=223.01850 \mathrm{u}$

Mass of ${ }_{82}^{209} \mathrm{~Pb}, m_{2}=208.98107 \mathrm{u}$

Mass of ${ }_{6}^{14} \mathrm{C}, m_{3}=14.00324 \mathrm{u}$

Hence, the Q-value of the reaction is given as:

Q = (m1 − m2 − m3) c2

= (223.01850 − 208.98107 − 14.00324) c2

= (0.03419 c2) u

But 1 u = 931.5 MeV/c2

∴Q = 0.03419 × 931.5

= 31.848 MeV

Hence, the Q-value of the nuclear reaction is 31.848 MeV. Since the value is positive, the reaction is energetically allowed.

Now take $a{ }_{2}^{4} \mathrm{He}$ emission nuclear reaction:

${ }_{88}^{223} \mathrm{Ra} \longrightarrow{ }_{86}^{229} \mathrm{Rn}+{ }_{2}^{4} \mathrm{He}$

We know that:

Mass of ${ }_{88}^{223} \mathrm{Ra}, m_{1}=223.01850$

Mass of ${ }_{82}^{219} \mathrm{Rn}, m_{2}=219.00948$

Mass of ${ }_{2}^{4} \mathrm{He}, m_{3}=4.00260$

Q-value of this nuclear reaction is given as:

Q = (m1 − m2 − m3) c2

= (223.01850 − 219.00948 − 4.00260) C2

= (0.00642 c2) u

= 0.00642 × 931.5 = 5.98 MeV

 

Hence, the Q value of the second nuclear reaction is 5.98 MeV. Since the value is positive, the reaction is energetically allowed.