Question

1) Americium-244 is a rare isotope of Americium. What is the mass defect of Americium-244?

Use the following values for atomic and neutron masses when calculating your answer:

$^{244}_{95} \text{Am:}\, 244.064279\,\mathrm{u}$
$^{1}_{1} \text{H:}\, 1.007825\,\mathrm{u}$
$\text{n:}\, 1.008665\,\mathrm{u}$

2)

Americium-244 undergoes beta (β-) decay. Select all of the products of this decay from the list below:

Select one or more:

$^{244}_{94} \text{Pu}$ ,

$^{0}_{+1}\mathrm{e}$ ,

$^{244}_{96} \text{Cm}$ ,

$^{0}_{-1}\mathrm{e}$ ,

$^{245}_{95} \text{Pu}$ ,

$^{244}_{95} \text{Am*}$ ,

$^{240}_{93} \text{Np}$ ,

3)

The half life of Americium-244 is 10.1 hours. After time 28.3 hours, how much of Americium-244 would be left in a sample that initially contained 10.8 g of Americium-244?

Answer 1

(1). Given, Atomicmass of Am, M 244.064279 u Mass of proton m 1.007825 u Mass ofneutron m. 1.008665 u Calculate the mass defect Here, Z is the atomic number and A is the mass number Substitute the values. 95x 1.0078254(244-95)×1.008665-244-064279 95.743375 +150-291085-244.064279 1.970181 u AM Hence, mass defect is 1.970181u

Let X and (e) are the products. Balance the atomic number, 95 Z+(-1) Z-96 Therefore, the product has an atomic number 96 and mass number 244. In the periodic table this element is Cm 244 96 Hence, the products are #Cm and Ole

Given Halflife ty2 -10.1 hours Total time 28.3 hours Initial amount N. -10.8 gm The number ofhalf lives, %2 28.3 10.1 2.8 Calculate the remaining amount of Americium. Substitute the values 2.8 N-10.8x = 10.8× 6.9644 -1.55 gm Hence, remaining amount of Americium is 1.55 gm