Question 13.23:
In a periodic table the average atomic mass of magnesium is given as 24.312 u. The average value is based on their relative natural abundance on earth. The three isotopes and their masses are (23.98504u),
(24.98584u) and (25.98259u). The natural abundance of is 78.99% by mass. Calculate the abundances of other two isotopes.
Answer:
Average atomic mass of magnesium, m = 24.312 u
Mass of magnesium isotope , m1 = 23.98504 u
Mass of magnesium isotope , m2 = 24.98584 u
Mass of magnesium isotope , m3 = 25.98259 u
Abundance of , η1= 78.99%
Abundance of , η2 = x%
Hence, abundance of , η3 = 100 − x − 78.99% = (21.01 − x)%
We have the relation for the average atomic mass as:
Hence, the abundance of is 9.3% and that of is 11.71%.
Question 13.24:
The neutron separation energy is defined as the energy required to remove a neutron from the nucleus. Obtain the neutron separation energies of the nuclei and
from the following data:
= 39.962591 u
) = 40.962278 u
= 25.986895 u
) = 26.981541 u
Answer:
For
For
A neutron is removed from a nucleus. The corresponding nuclear reaction can be written as:
It is given that:
Mass = 39.962591 u
Mass ) = 40.962278 u
Mass = 1.008665 u
The mass defect of this reaction is given as:
Δm =
∴Δm = 0.008978 × 931.5 MeV/c2
Hence, the energy required for neutron removal is calculated as:
For , the neutron removal reaction can be written as:
It is given that:
Mass = 26.981541 u
Mass = 25.986895 u
The mass defect of this reaction is given as:
Hence, the energy required for neutron removal is calculated as:
Question 13.25:
A source contains two phosphorous radio nuclides (T1/2 = 14.3d) and (T1/2 = 25.3d). Initially, 10% of the decays come from . How long one must wait until 90% do so?
Answer:
Half life of , T1/2 = 14.3 days
Half life of , T’1/2 = 25.3 days
nucleus decay is 10% of the total amount of decay.
The source has initially 10% of nucleus and 90% of nucleus.
Suppose after t days, the source has 10% of nucleus and 90% of nucleus.
Initially:
Number of nucleus = N
Number of nucleus = 9 N
Finally:
Number of
Number of
For nucleus, we can write the number ratio as:
For , we can write the number ratio as:
On dividing equation (1) by equation (2), we get:
Hence, it will take about 208.5 days for 90% decay of .
Question 13.26:
Under certain circumstances, a nucleus can decay by emitting a particle more massive than an α-particle. Consider the following decay processes:
Calculate the Q-values for these decays and determine that both are energetically allowed.
Answer:
Take a emission nuclear reaction:
We know that:
Mass of m1 = 223.01850 u
Mass of m2 = 208.98107 u
Mass of , m3 = 14.00324 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 emission nuclear reaction:
We know that:
Mass of m1 = 223.01850
Mass of m2 = 219.00948
Mass of , m3 = 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.