Question

Question: Prove that, to three-digit accuracy, \({\mathbf{h = 4}}{\mathbf{.14 \times 1}}{{\mathbf{0}}^{{\mathbf{ - 15}}}}{\mathbf{ eV}} \cdot {\mathbf{s}}\), as stated in the text.

Short answer

It is proved that to three-digit accuracy, \(h = 4.14 \times {10^{ - 34}}\;{\text{eV}} \cdot {\text{s}}\), as stated in the text.

Step-by-step solution

Concept Introduction

A sinusoidal wave's frequency is defined as the number of complete oscillations made by any wave constituent per unit of time. One may comprehend that if a body is in periodic motion, it has completed one cycle after passing through a series of events or locations and returning to its original condition using the notion of frequency. As a result, frequency is a quantity that describes the rate at which oscillations and vibrations occur.

Consider a photon of frequency \(f\). Hence the energy of the photon is –

\({{\text{E}}_{\text{\nu }}}{\text{ = hf}}\)

Evaluating value of \(h\)

Consider a photon of frequency \({\text{f}}\). Hence the energy of the photon is –

\({E_\nu } = hf\)

The Plank's constant \({\text{h}}\) is given by:

\(h = 6.62 \times {10^{ - 34}}{\text{\;J}} \cdot {\text{s}}\)

Hence, the energy of the photon in \({\text{J}}\) is:

\({E_\nu } = \left( {{\text{6}}.62 \times {{10}^{ - 34}}\;{\text{J}} \cdot {\text{s}}} \right)f\)

Now, it is known that –

\(1\;{\text{eV}} = 1.60 \times {10^{ - 19}}{\text{\;J}}\)

Hence, if the energy is converted in \({\text{eV}}\), it is obtained that –

\(\begin{gathered} {E_\nu } = \left( {6.62 \times {{10}^{ - 34}}\;{\text{Js}}} \right)\left( {\frac{{(1\;{\text{eV}})}}{{\left( {1.60 \times {{10}^{ - 19}}\;{\text{J}}} \right)}}} \right)f \\ = \left( {4.14 \times {{10}^{34}}\;{\text{eV}} \cdot {\text{s}}} \right){\text{f}} \\ = hf \\ \end{gathered} \)

Therefore, it is proved that \({\text{h = 4}}{\text{.14 \times 1}}{{\text{0}}^{{\text{ - 34}}}}{\text{eV}} \cdot {\text{s}}\).