You learned in Chapter 41 that the binding energy of the electron in a hydrogen atom is $13.6eV$.

a. By how much does the mass decrease when a hydrogen atom is formed from a proton and an electron? Give your answer both in atomic mass units and as a percentage of the mass of the hydrogen atom.

b. By how much does the mass decrease when a helium nucleus is formed from two protons and two neutrons? Give your answer both in atomic mass units and as a percentage of the mass of the helium nucleus.

c. Compare your answers to parts a and b. Why do you hear it said that mass is “lost” in nuclear reactions but not in chemical reactions?

Bending energy of a nucleus to eject the one nucleon from it is

$∆\mathrm{E}=∆{\mathrm{mc}}^2$

Here,$∆\mathrm{m}$Is mass difference and c is speed of the light

$$\begin{gathered} ∆\mathrm{m}=13.6\mathrm{eV} \\ =\left(13.6\mathrm{eV}\right)\left(\frac{1.0\mathrm{x}{10}^{-6}\mathrm{Mev}}{1.0\mathrm{ev}}\right) \\ ∆\mathrm{m}=\frac{13.6\mathrm{x}{10}^{-6}\mathrm{MeV}}{{\mathrm{c}}^2} \end{gathered}$$

The relation between atomic mass unit and $\mathrm{MeV}$is

$\left(1u\right)c^2=931.5MeV$

Rewrite this equation for 1 MeV

$1\mathrm{MeV}=\frac{\left(1\mathrm{u}\right){\mathrm{c}}^2}{931.5}$

substitute

$$\begin{gathered} \frac{\left(1\mathrm{u}\right){\mathrm{c}}^2}{931.5}\mathrm{for}1\mathrm{MeV}\mathrm{in}\mathrm{the}\mathrm{eqution}∆\mathrm{m}=\frac{13.6\mathrm{x}{10}^{-6}\mathrm{MeV}}{{\mathrm{C}}^2} \\ ∆\mathrm{m}=\frac{\left(13.6\mathrm{x}{10}^{-6}\right)\left({\displaystyle \frac{\left(1\mathrm{u}\right){\mathrm{c}}^2}{931.5}}\right)}{{\mathrm{c}}^2} \\ =1.46\mathrm{x}{10}^{-8}\mathrm{u} \end{gathered}$$

Therefore, the amount of mass decrease when the hydrogen formed from one electron and one proton is 1.46x10^-8u

Hydrogen atom consist one proton and one electron, so the total mass of the hydrogen atom is,

$m=m_c+m_p$

Here, ${\mathrm{m}}_{\mathrm{c}}$is mass of electron and ${\mathrm{m}}_{\mathrm{p}}$is mass of proton.

Substitute $0.00055\mathrm{u}$for m_c and $1.0073u$ for _{${\mathrm{m}}_{\mathrm{p}}$}.

$$\begin{gathered} \mathrm{m}=0.00055\mathrm{u}+1.0073\mathrm{u} \\ =1.00785\mathrm{u} \end{gathered}$$

Find the percentage of mass difference of hydrogen atom when it formed from one electron and

One proton can be calculated as follows.

$$\begin{gathered} \%=\frac{∆m}{m}x100 \\ =\left(\frac{1.46x{10}^{-8}u}{1.00785u}\right)x100 \\ =1.45x{10}^{-6} \end{gathered}$$

Therefore, the percentage of mass difference of hydrogen atom when it formed from one electron

and one proton is $1.45x10-6\%$

Mass difference of a nucleus is,

$∆m=Z{m}_p+\left(A-Z\right)m_n-M$

Here, Z is atomic number of the element of nucleus , A is mass number of the nucleus and M is atomic mass of the nucleus

Substitute 2 for Z, 4 for A, $1.0073\mathrm{u}$for ${\mathrm{m}}_{\mathrm{p}}$$1.0087\mathrm{u}$for ${\mathrm{m}}_{\mathrm{n}}$and $4.0026\mathrm{u}$for M and solve for the mass difference of helium nucleus,

$$\begin{gathered} ∆m=2(1.0073u)+(4-2)(1.0087u)-4.0026u \\ =0.0294u \\ =0.0304u \end{gathered}$$

Therefore, the mass decrease when helium nucleus formed from two protons and two neutrons is 0.$0.0304\mathrm{u}$

Find the percentage of mass difference of helium nucleus when it formed from two protons and two neutrons can be calculated as follows.$$\begin{gathered} \%=\frac{∆m}{m}x100 \\ =\left(\frac{0.0304u}{4.0026u}\right)x100 \\ =0.76 \end{gathered}$$

Therefore the percentage of mass difference of helium nucleus when it formed from two protons and two neutrons is $0.76\%$

The formation of hydrogen atom from one electron and one proton is a chemical reaction. The formation of helium nucleus from two protons and two neutrons is a nuclear reaction. From the above two results, in a chemical reaction the loss of mass is negligible. So mass is lost in nuclear reactions but not in chemical reactions.