Using the high-precision values of h,cand egiven on the text's inside front cover, show that the product hccan be expressed as$\mathbf{1240}\mathbf{}\mathbf{eV}\mathbf{·}\mathbf{nm}$

The relationship between the frequency of a photon and the energy of that photon is defined by Planck’s Constant. It explains the nature of particles and waves. It is denoted by h.

It is known that $h=6.62607004\times {10}^{-34}\mathrm{J}.\mathrm{s}$, and $c=2.99792458\times {10}^8\mathrm{m}/\mathrm{s}$.

Determine the value of hc by taking the product of h and c.

$$\begin{gathered} hc=6.62607004\times {10}^{-34}\mathrm{J}.\mathrm{s}\times 2.99792458\times {10}^8\mathrm{m}/\mathrm{s} \\ =1.98644582\times {10}^{-25}\mathrm{J}.\mathrm{s} \end{gathered}$$

Consider that the potential difference of 1V is applied on an electron.

Write the expression for the electric energy.

E = eV

Here, e is the charge on the electron with the value $1.60217662\times {10}^{-19}\mathrm{C}$.

Substitute all the values in the above expression.

$\begin{array}{l}\mathit{}E=1.60217662\times {10}^{-19}\mathrm{C}\times 1\mathrm{V}\\ =1.60217662\times {10}^{-19}\mathrm{V}·\mathrm{C}\\ 1\mathrm{eV}=1.60217662\times {10}^{-19}\mathrm{J}\end{array}$

Convert the units of hc from J to eV.

$$\begin{gathered} hc=1.98644582\times {10}^{-25}\mathrm{J}\times \frac{1\mathrm{eV}}{1.60217662\times {10}^{-19}\mathrm{J}}.\mathrm{m} \\ =1.98644582\times {10}^{-25}\times \frac{1\mathrm{eV}}{1.60217662\times {10}^{-19}\mathrm{J}}\times \frac{1\times {10}^9\mathrm{nm}}{1\mathrm{m}} \\ =1239.84197\mathrm{eV}·\mathrm{n}m \\ =1240\mathrm{eV}·\mathrm{nm} \end{gathered}$$

Thus, it is proved that the product of hc is $1240\mathrm{eV}·\mathrm{nm}$.