Question

The eyes of certain member of the reptile family pass a single visual signal to the brain when the visual receptors are stuck by photons of wavelength 662nm. 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. $(h=6.62\times {10}^{-34}J\times s)$

Answer 1

wavelength of photons $=662nm=662\times {10}^{-5}m$
speed of light (c) $=3\times {10}^{8}m/s$
frequency of photons $=\frac{c}{\lambda }=\frac{3\times {10}^{8}m/s}{662\times {10}^{-5}}=\frac{3\times {10}^{13}}{662}=4.53\times {10}^{10}{s}^{-1}$
Energy of 1 proton $=hv$
$=6.62\times {10}^{-34}Js\times 4.53\times {10}^{10}{s}^{-1}$
$=2.9988\times {10}^{-24}J$
Numbe of protons required $=\frac{Totalenergyrequired}{energyof1proton}$
$\frac{3.0\times {10}^{-14}}{2.9988\times {10}^{-24}}$
Since 209988 $\approx$ 3.0, we have
$\frac{{10}^{-14}}{{10}^{-24}}$
$={10}^{10}$ protons are needed.