Show that $\mathbf{ħ}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{06}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{34}}\mathbf{}\mathbf{J}\mathbf{.}\mathbf{s}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{59}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{16}\mathbf{}}\mathbf{eV}\mathbf{}\mathbf{.}\mathbf{s}\mathbf{}\mathbf{.}$.

The value of h (from Appendix B) is,

$h=6.62606876\times {10}^{-34}\mathrm{J}.\mathrm{s}$

The quantization of angular momentum is a modified form of Planck's constant known as h-bar, or the reduced Planck's constant, which equals h divided by 2.

Using the concept of energy per cycle and the value of Planck's constant, get the required value of the constant.

Formula:

The value of is given by,

$ħ=\raisebox{1ex}{$h$}\!\left/ \!\raisebox{-1ex}{$2\pi $}\right.$ …..(1)

One way to think of the units of h is that, because of the equation of the photon energy as given below.

E = hf

The fact that f in cycles/second, then the “explicit” units for h should be J .s/cycle .

Then, since $2\mathrm{\pi }\mathrm{rad}/\mathrm{cycle}$ in a conversion factor for$\mathrm{cycles}\to \mathrm{radians}$, equation (1) can be thought of as the Planck’s constant expressed in terms of radians instead of cycles.

Using the given data in equation (1), the value of is thus given by:

$$\begin{gathered} ħ=\frac{h}{2\pi } \\ =\frac{6.62606876\times {10}^{-34}\mathrm{J}.\mathrm{s}}{2\times 3.14} \\ =1.05457\times {10}^{-34}\mathrm{J}.\mathrm{s} \\ =\frac{1.05457\times {10}^{-34}\mathrm{J}.\mathrm{s}}{1.6021765\times {10}^{-19}\mathrm{J}/\mathrm{eV}} \\ \mathrm{ħ}=6.59\times {10}^{-16}\mathrm{eV}.\mathrm{s} \end{gathered}$$

Hence, the value of $ħ\mathrm{is}1.06\times {10}^{-34}\mathrm{J}.\mathrm{s}\mathrm{or}6.59\times {10}^{-16}\mathrm{eV}.\mathrm{s}$.