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Q8P

Expert-verifiedFound in: Page 1302

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**a. Show that the mass M of an atom is given approximately by ${{\mathit{M}}}_{\mathbf{a}\mathbf{p}\mathbf{p}}{\mathbf{=}}{\mathit{A}}{{\mathit{m}}}_{{\mathbf{p}}}$, where A is 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?**

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

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

**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)

The mass number A* *is 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:

${M}_{app}=p{m}_{p}+n{m}_{n}\phantom{\rule{0ex}{0ex}}=\left(p+n\right){m}_{p}\phantom{\rule{0ex}{0ex}}=A{m}_{p}$

Also, the mass of the electrons is* *negligible 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.

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

${\mathrm{M}}_{\mathrm{app}}=\left(1\right)\left(1.007276\mathrm{u}\right)\phantom{\rule{0ex}{0ex}}=1.007276\mathrm{u}\left(\because {\mathrm{m}}_{\mathrm{p}}=1.007276\mathrm{u}\right)$

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:

$\mathrm{percentage}\mathrm{deviation}=\frac{1.007276\mathrm{u}-1.007825\mathrm{u}}{1.007825\mathrm{u}}\times 100\phantom{\rule{0ex}{0ex}}=-0.054\%\phantom{\rule{0ex}{0ex}}\approx -0.05\%$

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

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

${\mathrm{M}}_{\mathrm{app}}=\left(1\right)\left(1.007276\mathrm{u}\right)\phantom{\rule{0ex}{0ex}}=31.225556\mathrm{u}\left(\because {\mathrm{m}}_{\mathrm{p}}=1.007276\mathrm{u}\right)$

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:

$\mathrm{percentage}\mathrm{deviation}=\frac{31.225556-30.973762\mathrm{u}}{30.973762\mathrm{u}}\times 100\phantom{\rule{0ex}{0ex}}=0.81\%$

Hence, the value of the deviation is 0.81% .

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

${\mathrm{M}}_{\mathrm{app}}=\left(120\right)\left(1.007276\mathrm{u}\right)\left(\because {\mathrm{m}}_{\mathrm{p}}=1.007276\mathrm{u}\right)\phantom{\rule{0ex}{0ex}}=120.87312\mathrm{u}$

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:

$\mathrm{percentage}\mathrm{deviation}=\frac{120.87312u-119.902197\mathrm{u}}{119.902197\mathrm{u}}\times 100\phantom{\rule{0ex}{0ex}}=0.81\%$

Hence, the value of the deviation is 0.81% .

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

${\mathrm{M}}_{\mathrm{app}}=\left(197\right)\left(1.007276\mathrm{u}\right)\left(\because {\mathrm{m}}_{\mathrm{p}}=1.007276\mathrm{u}\right)\phantom{\rule{0ex}{0ex}}=198.433372\mathrm{u}$

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:

$\mathrm{percentage}\mathrm{deviation}=\frac{198.433372\mathrm{u}-196.966552\mathrm{u}}{196.966552\mathrm{u}}\times 100\phantom{\rule{0ex}{0ex}}=0.74\%$

Hence, the value of the deviation is 0.74% .

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

${\mathrm{M}}_{\mathrm{app}}=\left(239\right)\left(1.007276\mathrm{u}\right)\left(\because {\mathrm{m}}_{\mathrm{p}}=1.007276\mathrm{u}\right)\phantom{\rule{0ex}{0ex}}=240.738964\mathrm{u}$

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:

$\mathrm{percentage}\mathrm{deviation}=\frac{240.738964\mathrm{u}-239.052157\mathrm{u}}{239.052157\mathrm{u}}\times 100\phantom{\rule{0ex}{0ex}}=0.71\%$

Hence, the value of the deviation is 0.71 %.

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.

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