Question

The eyes of a certain reptile family pass a single visual signal to the brain when the visual receptors are struck by photons of wavelength 662 nm.If a total energy of $3.0\times {10}^{-14}J$ is required to trip the signal,what is the minimum number of photons that must strike the receptor.($h=6.62\times {10}^{-34}Js$)

Answer 1

The energy of photon $=hc/\lambda =\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{6.62\times {10}^{-9}}$

$=3\times {10}^{-19}$

The number of photons required is $=\frac{3\times {10}^{-14}J}{3\times {10}^{-19}}={10}^5$