Question

The mass of a deuterium nucleus 21H is less than its components masses. Calculate the mass defect.________ amu

Answer 1

Answer:

The mass defect of a deuterium nucleus is 0.001848 amu.

Explanation:

The deuterium is:

$^{A}_{Z}X \rightarrow ^{2}_{1}H$  

The mass defect can be calculated by using the following equation:

$\Delta m = [Zm_{p} + (A - Z)m_{n}] - m_{a}$

Where:

Z: is the number of protons = 1

A: is the mass number = 2      

$m_{p}$: is the proton's mass = 1.00728 amu  

$m_{n}$: is the neutron's mass = 1.00867 amu

$m_{a}$: is the mass of deuterium = 2.01410178 amu

Then, the mass defect is:

$\Delta m = [1.00728 amu + (2- 1)1.00867 amu] - 2.01410178 amu = 0.001848 amu$

Therefore, the mass defect of a deuterium nucleus is 0.001848 amu.

I hope it helps you!  

Answer 2

Answer:

0.002389

Explanation:

Use the equation the other person used to answer but with the numbers that are given on this assignment!