Question

Confirm the statement in the text that the range of photon energies for visible light is \({\rm{1}}{\rm{.63}}\) to \({\rm{3}}{\rm{.26 eV}}\), given that the range of visible wavelengths is \({\rm{380}}\) to \({\rm{760 nm}}\).

Short answer

It is confirmed that the energy of the visible light photon varies from \({\rm{1}}{\rm{.63 eV}}\) to\({\rm{3}}{\rm{.26 eV}}\) when the range of visible wavelengths is \({\rm{380 - 760 nm}}\).

Step-by-step solution

Concept Introduction

The distance between the identical places between two successive waves is determined when the wavelength of a wave is measured.

Information Provided

• Range of visible wavelength is:\(380 - 760\,{\rm{nm}}\).
• Range of photon energy should be: \(1.63 - 3.26\,{\rm{eV}}\).

Range of energy for photon

The energy of a photon is given by –

\(E = \frac{{hc}}{\lambda }\)

When

\(\begin{aligned}{}\lambda = 380\,{\rm{nm}}\\ = 380 \times {10^{ - 9}}\,{\rm{m}}\end{aligned}\).

The energy of the photon is –

\(\begin{aligned}{}E &= \frac{{\left( {4.13 \times {{10}^{ - 15}}\,{\rm{eV}}{\rm{.s}}} \right)\left( {3 \times {{10}^8}\,{\rm{m/s}}} \right)}}{{\left( {380 \times {{10}^{ - 9}}\,{\rm{m}}} \right)}}\\ &= 3.26\,{\rm{eV}}\end{aligned}\)

When the wavelength of the photon is

\(\begin{aligned}{}\lambda &= 760\,{\rm{nm}}\\ &= 760 \times {10^{ - 9}}\,{\rm{m}}\end{aligned}\).

The energy of the photon is –

\(\begin{aligned}{}E &= \frac{{\left( {4.13 \times {{10}^{ - 15}}\,{\rm{eV}}{\rm{.s}}} \right)\left( {3 \times {{10}^8}\,{\rm{m/s}}} \right)}}{{\left( {760 \times {{10}^{ - 9}}\,{\rm{m}}} \right)}}\\ &= 1.63\,{\rm{eV}}\end{aligned}\)

Therefore, the value for energies are obtained as \(3.26\,{\rm{eV}}\)and\(1.63\,{\rm{eV}}\).