Question

The mass defect in ${}_{2}H{e}^3$ is $\underset{–}{}$ , if m${}_p=$1.00727 amu, $m_n=$1.008665 amu, mass of ${}_{2}H{e}^3=$3.01664 amu

• A. 0.00657 amu
• B. 0.00621 amu
• C. 0.0657 amu
• D. 0.6877 amu

Answer 1

Answer: (A)

0.00657 amu

The difference between atomic weight and mass number i.e. mass of elementary particle in the nucleus is known as mass defect. Since, ${}_3^2\text{He}$ has two protons(atomic number = proton) and one neutron(mass number = proton + neutron).
Hence, Mass defect is
$\Delta m=\left[2\right(m_p)+1(m_n\left)\right]-M_{He}$
where, variables have their usual meanings.
$\Delta m=\left[2\right(1.00727)+1(1.008665\left)\right]-3.01664$
$\Delta m=[2.01454+1.008665]-3.01664$
$\Delta m=3.023205-3.01664$
$\Delta m=0.006565amu$
$\Delta m\approx 0.00657amu$