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### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# Under ideal conditions, a visual sensation can occur in the human visual system if the light of wavelength 550nm is absorbed by the eye’s retina at a rate as low as 100 photons per second. What is the corresponding rate at which energy is absorbed by the retina?

The rate at which energy is absorbed by photon is $3.6×{10}^{-17}\mathrm{W}.$

See the step by step solution

## Describe the expression of the energy of the photon:

The energy of a photon is given by,

${E}{=}\frac{hc}{\lambda }$ ….. (1)

Here, h is Planck’s constant, ${\lambda }$ is the wavelength, and c is the speed of light.

## Determine the energy absorbed by the retina:

Consider the given data as below.

The wavelength, $\lambda =550nm.$

Plank’s constant, $\mathrm{h}=6.626×{10}^{-34}\mathrm{J}.\mathrm{s}.$

The speed of light, $\mathrm{c}=3×{10}^{8}\mathrm{m}/\mathrm{s}$

Substitute the below values in eq 1.

$\mathrm{c}=3×{10}^{8}\mathrm{m}/\mathrm{s}\phantom{\rule{0ex}{0ex}}\mathrm{h}=6.626×{10}^{-34}\mathrm{J}·\mathrm{s}\phantom{\rule{0ex}{0ex}}\mathrm{\lambda }=550\mathrm{nm}$

Find the energy absorbed by retina as follows.

$\mathrm{energyabsorbedbyretina}=\left(\mathrm{photon}\mathrm{per}\mathrm{second}\right)\left(\mathrm{energy}\mathrm{of}\mathrm{the}\mathrm{photon}\right)\phantom{\rule{0ex}{0ex}}=\left(100\mathrm{photons}/\mathrm{s}\right)\left(3.6×{10}^{-19}\mathrm{J}/\mathrm{photons}\right)\phantom{\rule{0ex}{0ex}}=3.6×{10}^{-17}\mathrm{J}/\mathrm{s}\phantom{\rule{0ex}{0ex}}=3.6×{10}^{-17}\mathrm{W}$

Hence, the rate at which energy is absorbed by photon is $3.6×{10}^{-17}\mathrm{W}.$