Question

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

Short answer

It is proved that the product of hc is 1240 eV.nm.

Step-by-step solution

Significance of Planck’s Constant

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.

Determination of the value of  

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.

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

Determination of the value of hc

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 $1240\mathrm{eV}·\mathrm{nm}$.