Question

a. Show that the massMof an atom is given approximately by ${\mathit{M}}_{\mathbf{a}\mathbf{p}\mathbf{p}}\mathbf{=}\mathit{A}{\mathit{m}}_{\mathbf{p}}$, whereAis the mass number and is the proton mass. For (b) ${}^{\mathbf{1}}\mathbf{H}$, (c)${}^{\mathbf{31}}\mathbf{P}$,(d)${}^{\mathbf{120}}\mathbf{Sn}$, (e) ${}^{\mathbf{197}}\mathbf{Au}$, and (f) ${}^{\mathbf{239}}\mathbf{Pu}$, use Table 42-1 to find the percentage deviation between ${\mathbf{M}}_{\mathbf{app}}$and $\mathbf{M}$:

role="math" localid="1662047222746" $\mathbf{percentage}\mathbf{}\mathbf{deviation}\mathbf{=}\frac{{\mathbf{M}}_{\mathbf{app}}\mathbf{-}\mathbf{M}}{\mathbf{M}}\mathbf{\times }\mathbf{100}$

(g) Is a value of${\mathbf{M}}_{\mathbf{app}}$accurate enough to be used in a calculation of a nuclear binding energy?

Short answer

1. The mass of an atom is given by $M_{app}=A{m}_p$.
2. The percentage deviation of${}^1\mathrm{H}$is -0.05 %.
3. The percentage deviation of ${}^{31}\mathrm{P}$is 0.81%.
4. The percentage deviation of ${}^{120}\mathrm{Sn}$is 0.81%.
5. The percentage deviation of ${}^{197}\mathrm{Au}$is 0.74% .
6. The percentage deviation of ${}^{239}\mathrm{Pu}$is 0.71% .
7. No, there is no value of accurate enough to be used in a calculation of a nuclear binding energy.

Step-by-step solution

Write the given data

1. Given particles:${}^1\mathrm{H},{}^{31}\mathrm{P},{}^{120}\mathrm{Sn},{}^{197}\mathrm{Au},{}^{239}\mathrm{Pu}$
2. The formula of percentage deviation,localid="1662047170223" $=\frac{M_{app}-M}{M}\times 100$

Determine the concept of mass  

The nucleus of an atom is made up of protons and neutrons that together are called nucleons. Thus, the mass of an atom can be determined on that basis as the mass number is defined as the sum of the number of nucleons. Now, using the given percentage mass deviation formula, we can get the required values of the deviations of the element.

Formulae:

The mass number of an element as follows:

A = n + p …… (i)

The percentage deviation from the mass is as follows:

$percetagedeviation=\frac{M_{app}-M}{M}\times 100$ …… (ii)

a) Determine the mass of an atom

The mass number Ais the number of nucleons in an atomic nucleus. Since $m_p\approx m_n$, the mass of the nucleus is approximately is given using equation (i) as follows:

$$\begin{gathered} M_{app}=p{m}_p+n{m}_n \\ =\left(p+n\right)m_p \\ =A{m}_p \end{gathered}$$

Also, the mass of the electrons isnegligible since it is much less than that of the nucleus.

Hence, the mass of an atom is given by$M_{app}=A{m}_p$ , where is the mass of the proton.

b) Calculate the percentage deviation H1

From the calculations of part (a), we get the apparent mass from equation for as follows:

$$\begin{gathered} {\mathrm{M}}_{\mathrm{app}}=\left(1\right)\left(1.007276\mathrm{u}\right) \\ =1.007276\mathrm{u}\left(\because {\mathrm{m}}_{\mathrm{p}}=1.007276\mathrm{u}\right) \end{gathered}$$

The actual mass of ${}^1\mathrm{H}$is M = 1.007825u (from Table 42-1).

Thus, the percentage deviation is given using the above data in equation (ii) as follows:

$$\begin{gathered} \mathrm{percentage}\mathrm{deviation}=\frac{1.007276\mathrm{u}-1.007825\mathrm{u}}{1.007825\mathrm{u}}\times 100 \\ =-0.054\% \\ \approx -0.05\% \end{gathered}$$

Hence, the value of the deviation is -0.05% .

c) Calculate the percentage deviation P31

From the calculations of part (a), determine apparent mass from equation (a) as follows:

$$\begin{gathered} {\mathrm{M}}_{\mathrm{app}}=\left(1\right)\left(1.007276\mathrm{u}\right) \\ =31.225556\mathrm{u}\left(\because {\mathrm{m}}_{\mathrm{p}}=1.007276\mathrm{u}\right) \end{gathered}$$

The actual mass of ${}^{31}\mathrm{P}$is$\mathrm{M}=30.973762\mathrm{u}$ (from Table 42-1).

Thus, the percentage deviation is given using the above data in equation (ii) as follows:

$$\begin{gathered} \mathrm{percentage}\mathrm{deviation}=\frac{31.225556-30.973762\mathrm{u}}{30.973762\mathrm{u}}\times 100 \\ =0.81\% \end{gathered}$$

Hence, the value of the deviation is 0.81% .

d) Calculate the percentage deviation Sn120

From the calculations of part (a), we get the apparent mass from equation (a) as follows:

$$\begin{gathered} {\mathrm{M}}_{\mathrm{app}}=\left(120\right)\left(1.007276\mathrm{u}\right)\left(\because {\mathrm{m}}_{\mathrm{p}}=1.007276\mathrm{u}\right) \\ =120.87312\mathrm{u} \end{gathered}$$

The actual mass of ${}^{120}\mathrm{Sn}$is$M=119.902197u$ (from Table 42-1).

Thus, the percentage deviation is given using the above data in equation (ii) as follows:

$$\begin{gathered} \mathrm{percentage}\mathrm{deviation}=\frac{120.87312u-119.902197\mathrm{u}}{119.902197\mathrm{u}}\times 100 \\ =0.81\% \end{gathered}$$

Hence, the value of the deviation is 0.81% .

e) Calculate the percentage deviation Au197

From the calculations of part (a), we get the apparent mass from equation as follows:

$$\begin{gathered} {\mathrm{M}}_{\mathrm{app}}=\left(197\right)\left(1.007276\mathrm{u}\right)\left(\because {\mathrm{m}}_{\mathrm{p}}=1.007276\mathrm{u}\right) \\ =198.433372\mathrm{u} \end{gathered}$$

The actual mass of${}^{197}\mathrm{Au}$is M = 196.96652u (from Table 42-1).

Thus, the percentage deviation is given using the above data in equation (ii) as follows:

$$\begin{gathered} \mathrm{percentage}\mathrm{deviation}=\frac{198.433372\mathrm{u}-196.966552\mathrm{u}}{196.966552\mathrm{u}}\times 100 \\ =0.74\% \end{gathered}$$

Hence, the value of the deviation is 0.74% .

f) Calculate the percentage deviation Pu239

From the calculations of part (a), we get the apparent mass from equation as follows:

$$\begin{gathered} {\mathrm{M}}_{\mathrm{app}}=\left(239\right)\left(1.007276\mathrm{u}\right)\left(\because {\mathrm{m}}_{\mathrm{p}}=1.007276\mathrm{u}\right) \\ =240.738964\mathrm{u} \end{gathered}$$

The actual mass of ${}^{239}\mathrm{Pu}$is M = 230.052157u (from Table 42-1).

Thus, the percentage deviation is given using the above data in equation (ii) as follows:

$$\begin{gathered} \mathrm{percentage}\mathrm{deviation}=\frac{240.738964\mathrm{u}-239.052157\mathrm{u}}{239.052157\mathrm{u}}\times 100 \\ =0.71\% \end{gathered}$$

Hence, the value of the deviation is 0.71 %.

g) Calculate the value of the apparent mass that can be used for nuclear binding energy calculations

No. In a typical nucleus the binding energy per nucleon is several MeV, which is a bit less than 1% of the nucleon mass times $c^2$. This is comparable with the percent error calculated in parts (b) – (f), so we need to use a more accurate method to calculate the nuclear mass.