The eye can detect 5 times 10^4 photons per s | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The energy of a single photon of green light is given by $E = \frac{hc}{\lambda}$.Using $hc \approx 12400 \ 	ext{eV} \ \mathring{A}$, we get $E =

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(A) The energy of a single photon of green light is given by $E = \frac{hc}{\lambda}$.Using $hc \approx 12400 \ 	ext{eV} \ \mathring{A}$, we get $E = \frac{12400}{5000} \ 	ext{eV} = 2.48 \ 	ext{eV}$.Converting this to Joules: $E = 2.48 	imes 1.6 	imes 10^{-19} \ J \approx 3.97 	imes 10^{-19} \ J \approx 4 	imes 10^{-19} \ J$.The minimum intensity detected by the eye $(I_{Eye})$ is the product of photon flux and energy per photon:$I_{Eye} = (5 	imes 10^4 \ 	ext{photons}/m^2s) 	imes (4 	imes 10^{-19} \ J) = 2 	imes 10^{-14} \ W/m^2$.The sensitivity of a detector is inversely proportional to the minimum intensity it can detect.Therefore, the ratio of sensitivity is $\frac{S_{Eye}}{S_{Ear}} = \frac{I_{Ear}}{I_{Eye}} = \frac{10^{-13}}{2 	imes 10^{-14}} = \frac{10}{2} = 5$.Thus, the eye is $5$ times more sensitive than the ear.

The eye can detect $5 \times 10^4$ photons per square metre per second of green light $(\lambda = 5000 \ \mathring{A})$ while the ear can detect $10^{-13} \ W/m^2$. The factor by which the eye is more sensitive as a power detector than the ear is close to:

Similar Questions

$A$ parallel beam of light of wavelength $900 \, nm$ and intensity $100 \, W/m^2$ is incident on a surface perpendicular to the beam. The number of photons crossing $1 \, cm^2$ area perpendicular to the beam in one second is:

$A$ $200\, W$ sodium street lamp emits yellow light of wavelength $0.6\, \mu m$. Assuming it to be $25\%$ efficient in converting electrical energy to light, the number of photons of yellow light it emits per second is:

The fact that photons carry energy was established by

The momentum of a photon having energy $hv$ is ............

An electron-positron pair is created by $\gamma$-rays. What is the minimum energy of the $\gamma$-rays required for this process in $MeV$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module