Question

An alpha particle $({}^{4}He)$ has a mass of 4.00300 amu. A proton has a mass of 1.00783 amu and a neutron has a mass of 1.00867 amu respectively. The binding energy of alpha particle estimated from these data is the closest to
A. 27.9 MeV
B. 22.3 MeV
C. 35.0 MeV
D. 20.4 MeV

Answer 1

Hint: Study about an alpha particle. Try to know its constituent particle and then try to calculate the mass defect of the alpha particle. Then we can use the formula to calculate the binding energy.

Formula used:
Now, the binding energy of a particle is given by,
$B.E.=\Delta m{{c}^{2}}$

Complete Step-by-Step solution:
An alpha particle also commonly known as alpha ray $\left( \alpha \text{-ray} \right)$is a positively charged particle which is emitted by certain radioactive material. The alpha particle consists of 2 protons and 2 neutrons. So, it is similar to the ${}_{2}^{4}He$ nucleus.
Since, the alpha particle is represented by ${}_{2}^{4}He$ nucleus,
Number of protons in alpha particle is 2
And number of neutrons in alpha particle is 2
The binding energy can be defined as the minimum energy required to break a nucleus into its constituent protons and neutrons.
Now, to calculate the binding energy we have to first calculate the mass defect of alpha particle.
Given,
Mass of proton, ${{m}_{p}}=1.00783amu$
Mass of neutron, ${{m}_{n}}=1.00867amu$
Mass of ${}_{2}^{4}He$ =$4.00300amu$
So, mass defect $\Delta m$ of alpha particle is given by,
$\begin{align}
  & \Delta m=2{{m}_{p}}+2{{m}_{n}}-{{m}_{{}^{4}He}} \\
 & \Delta m=2\times 1.00783amu+2\times 1.00867amu-4.00300amu \\
 & \Delta m=2.01566amu+2.01734amu-4.00300amu \\
 & \Delta m=0.03amu \\
\end{align}$
Now, the binding energy of alpha particle is given by,
$\begin{align}
  & B.E.=\Delta m{{c}^{2}} \\
 & B.E.=0.03\text{ amu }{{\text{c}}^{\text{2}}} \\
\end{align}$
Here, amu means atomic mass unit.
$1amu=1.66\times {{10}^{-27}}Kg$
So, binding energy will be,
$\begin{align}
  & B.E.=0.03\times 1.66\times {{10}^{-27}}\text{Kg}\times {{\left( 3\times {{10}^{8}}m{{s}^{-1}} \right)}^{2}} \\
 & B.E.=4.482\times {{10}^{-12}}Kg{{m}^{2}}{{s}^{-2}} \\
 & B.E.=4.482\times {{10}^{-12}}Joule \\
\end{align}$
Converting Joule to eV,
$\begin{align}
  & B.E.=\dfrac{4.482\times {{10}^{-12}}}{1.67\times {{10}^{-19}}}eV \\
 & B.E.=26.8\times {{10}^{6}}eV \\
 & B.E.=26.8MeV \\
\end{align}$
Approximately we can take it as 27.9 MeV.
The correct option is (A).

Note: We can also calculate the binding energy by taking $1\text{ amu }{{\text{c}}^{2}}=931.5Me{{V}^{{}}}$.
So, now the binding energy will be,
$\begin{align}
  & B.E.=0.03\text{ amu }{{\text{c}}^{\text{2}}} \\
 & B.E.=0.03\times 931.5MeV \\
 & B.E.=27.945MeV \\
\end{align}$