Question

What is the mass excess ${\mathbf{∆}}_{\mathbf{1}}$of${\mathbf{}}^{\mathbf{1}}\mathbf{H}$(actual mass is 1.007825 u) in (a) atomic mass units and (b) $\mathbf{MeV}\mathbf{/}{\mathbf{c}}^{\mathbf{2}}$? What is the mass excess ${\mathbf{∆}}_{\mathit{n}}$of a neutron (actual mass is 1.008665u ) in (c) atomic mass units and (d) $\mathbf{MeV}\mathbf{/}{\mathbf{c}}^{\mathbf{2}}$? What is the mass excess ${\mathbf{∆}}_{\mathbf{120}}$of${\mathbf{}}^{\mathbf{120}}\mathbf{Sn}$(actual mass is119.902197 u) in (e) atomic mass units and (f) $\mathbf{MeV}\mathbf{/}{\mathbf{c}}^{\mathbf{2}}$?

Short answer

1. The mass excess${∆}_1$of${}^1\mathrm{H}$in atomic mass units is 0.007825u.
2. The mass excess ${∆}_1$of ${}^1\mathrm{H}$in $\mathrm{MeV}/{\mathrm{c}}^2$is data-custom-editor="chemistry" $\mathrm{MeV}/{\mathrm{c}}^2$.
3. The mass excess ${∆}_n$ofa neutron in atomic mass units is 0.008685 u.
4. The mass excess ${∆}_n$of a neutron in $\mathrm{MeV}/{\mathrm{c}}^2$is $+8.071\mathrm{MeV}/{\mathrm{c}}^2$.
5. The mass excess${∆}_{120}$of${}^{120}\mathrm{Sn}$ in atomic mass units is -0.09780 u.
6. The mass excess${∆}_{120}$of ${}^{120}\mathrm{Sn}$in$\mathrm{MeV}/{\mathrm{c}}^2$ is -91.10$\mathrm{MeV}/{\mathrm{c}}^2$.

Step-by-step solution

Given data

Actual mass of ${}^1\mathrm{H}$is ${\mathrm{m}}_{\mathrm{H}}=1.007825\mathrm{u}$.

Actual mass of a neutron is ${\mathrm{m}}_{\mathrm{n}}=1.008665\mathrm{u}$.

Actual mass of ${}^{120}\mathrm{Sn}$is${\mathrm{m}}_{\mathrm{n}}=119.902197\mathrm{u}$ .

Understanding the concept of mass excess  

Mass defect is the difference between the predicted mass and the actual mass of an atom's nucleus. It may sometimes be called mass excess. The mass excess of a nuclide is the difference between its actual mass and its mass number in atomic mass units. Now, using the calculated mass excess for each given nuclide, we can convert it into the required unit of MeV/c².

Formulae:

The mass excess of a nuclide,$\mathrm{Mass}\mathrm{excess}\left(\mathrm{D}\right)=\mathrm{Actual}\mathrm{mass}-\mathrm{Mass}\mathrm{in}\mathrm{amu}..........\left(1\right)$

The conversion of atomic mass unit into$\mathrm{MeV}/{\mathrm{c}}^2,1\mathrm{amu}=931.5\mathrm{Mev}/{\mathrm{c}}^2.................\left(2\right)$ ,

a) Calculation of the mass excess of  1H in atomic mass unit

Using the given data in equation (1), the mass excess of in atomic mass units is given as follows:

$$\begin{gathered} {∆}_1=1.007825\mathrm{u}-1.000000\mathrm{u} \\ =0.007825\mathrm{u} \end{gathered}$$

Hence, the value of the mass excess is 0.007825 u.

b) Calculation of the mass excess of  1H in MeV/c2

Now, using the result of part (a) and equation (2), we can get the mass excess of in$\mathrm{MeV}/{\mathrm{c}}^2$ as follows:

$$\begin{gathered} {∆}_1=0.007825\mathrm{u}\times \left(\frac{\left(931.5\mathrm{MeV}/{\mathrm{c}}^2\right)}{1\mathrm{u}}\right) \\ =+7.290\mathrm{MeV}/{\mathrm{c}}^2 \end{gathered}$$

Hence, the value of the mass excess is +7.290 MeV/${\mathrm{c}}^2$ .

c) Calculation of the mass excess of a neutron in atomic mass unit

Using the given data in equation (1), the mass excess${∆}_{\mathrm{n}}$of a neutron in atomic mass units is given as follows:

$$\begin{gathered} {∆}_{\mathrm{n}}=1.008665\mathrm{u}-1.000000\mathrm{u} \\ =0.008665\mathrm{u} \end{gathered}$$

Hence, the value of the mass excess is 0.008665 u.

d) Calculation of the mass excess of a neutron in MeV/c2

Now, using the result of part (a) and equation (ii), we can get the mass excess${∆}_{\mathrm{n}}$ of a neutron in as follows:

$$\begin{gathered} {∆}_{\mathrm{n}}=0.008665\mathrm{u}\times \left(\frac{\left(931.5\mathrm{MeV}/{\mathrm{c}}^2\right)}{1\mathrm{u}}\right) \\ =+8.071\mathrm{MeV}/{\mathrm{c}}^2 \end{gathered}$$

Hence, the value of the mass excess is $+8.071\mathrm{MeV}/{\mathrm{c}}^2$.

e) Calculation of the mass excess of  120Sn in atomic mass unit

Using the given data in equation (1), the mass excess ${∆}_{\mathrm{Sn}}$of ${}^{120}\mathrm{Sn}$in atomic mass units is given as follows:

$$\begin{gathered} {∆}_{\mathrm{Sn}}=119.902197\mathrm{u}-120.000000\mathrm{u} \\ =-0.09780\mathrm{u} \end{gathered}$$

Hence, the value of the mass excess is -0.9780 u.

 Step 8: f) Calculation of the mass excess of a neutron in MeV/c2

Now, using the result of part (a) and equation (2), we can get the mass excess${∆}_{\mathrm{Sn}}$ of${}^{120}\mathrm{Sn}$ in$\mathrm{MeV}/{\mathrm{c}}^2$ as follows:

$$\begin{gathered} {∆}_{\mathrm{n}}=-0.09780\mathrm{u}\times \left(\frac{\left(931.5\mathrm{MeV}/{\mathrm{c}}^2\right)}{1\mathrm{u}}\right) \\ =-91.10\mathrm{MeV}/{\mathrm{c}}^2 \end{gathered}$$

Hence, the value of the mass excess is$-91.10\mathrm{MeV}/{\mathrm{c}}^2$ .