Question

What is the energy of visible light ?

Answer 1

Hint: To solve this problem learn about visible light. Learn about the frequency range of visible light and use the energy relation for a wave to find the energy of visible light. Visible light is that section of electromagnetic waves that is perceivable to human eyes. Anything that glows or produces light has waves of visible light.

Formula used:
The energy of a wave is given by, \[E = h\nu \]
where \[E\] is the energy of the electromagnetic wave, \[\nu \] is the frequency of the wave and \[h\] is Planck's constant. The value of Planck’s constant is \[h = 6.62 \times {10^{ - 34}}JH{z^{ - 1}}\].

Complete step by step answer:
We know that the visible spectrum of light is a field of the electromagnetic spectrum that is visible to the human eye. The wavelength of the visible spectrum of light lies in the range of \[380\,nm\] to \[740\,nm\]. The frequency range of the visible spectrum of light lies in the range of \[405\,THz\] to \[790\,THz\].

Now, the energy of a wave is given by, \[E = h\nu \] .The value of Planck’s constant is \[h = 6.62 \times {10^{ - 34}}JH{z^{ - 1}}\].So, putting the minimum and maximum value of frequency in the formula we can get the range of energy of the visible spectrum of light.So, putting \[\nu = 405\,THz\] we get the minimum energy of visible light as,
\[{E_{\min }} = 6.62 \times {10^{ - 34}} \times 405 \times {10^{12}}J\]

Up on simplifying we have, \[{E_{\min }} = 2.681 \times {10^{ - 19}}J\]
Now, we also know, \[1eV = 1.602 \times {10^{ - 19}}J\]
So, in \[eV\] it will be, \[{E_{\min }} = 1.67eV\]
Now, putting \[\nu = 790THz\] we get the minimum energy of visible light as,
\[{E_{\max }} = 6.62 \times {10^{ - 34}} \times 790 \times {10^{12}}J\]
Up on simplifying we have, \[{E_{\max }} = 5.229 \times {10^{ - 19}}J\]
So, in \[eV\] it will be, \[{E_{\max }} = 3.22eV\].

Hence, the energy of visible light lies in the range of \[1.67\,eV - 3.22\,eV\].

Note: To find the energy of visible light one has to remember the range of either the wavelength or the frequency. From wavelength we can easily find the frequency using the relation \[c = \lambda \nu \] where \[c\] is the velocity of light in vacuum and \[\lambda \] is the wavelength of the respective wave. The energy of each colour of light in the visible spectrum has individual range e.g. for violet light it has a range \[(2.95 - 3.22)\,eV\] the red light lies in the range of \[(1.65 - 2)\,eV\]