Question

A 60 W bulb is placed at a distance of 4m from you. The bulb is emitting light of wavelength 600 nm uniformly in all directions. In 0.1 seconds, how many photons can enter your eye if the pupil of the eye is having a diameter of 2 mm?
A) $2.84 \times {10^{12}}$
B) $2.84 \times {10^{11}}$
C) $9.37 \times {10^{11}}$
D) $6.48 \times {10^{11}}$

Answer 1

The number of photons entering can be calculated using their intensity and energy in the photoelectric effect. After getting a number of photons entering in 1 second, we can calculate for them entering in 0.1 seconds.
Intensity (I) = $\dfrac{{Power}}{{Area}}$
Energy (E) = $\dfrac{{hc}}{\lambda }$
Planck’s constant (h) = $6.6 \times {10^{ - 34}}$
c = speed of light = $3 \times {10^8}m/s \\\ \\\$
For conversion:
1 m = ${10^9}nm$
1 m = 1000 mm

Complete step by step answer: Given:
Power (P) = 60 W
Wavelength $(\lambda )$ = 600 nm = $600 \times {10^{ - 9}}m$ [As 1 m = ${10^9}nm$]
Time (t) = 0.1 s
Now,
Intensity is defined as power per unit surface area and it can be given as:
Intensity (I) = $\dfrac{{Power}}{{Area}}$
Area = $4\pi {r^2}$ [We will use area of sphere (for bulb)]
r = distance from bulb to the person
r = 4 cm
Substituting the values:
$I = \dfrac{{60}}{{4 \times \pi \times {{\left( 4 \right)}^2}}}$ $W/{m^2}$
Energy that will enter the eye per second:
E = Intensity X area of eye in contact (circular)
Area in contact = $\pi {r^2}$
Since the diameter (d) of eye is given instead of radius (r) , we can use
$r = \dfrac{d}{2}$ [ As diameter = 2 X radius]

$E = I \times \pi {r^2} \\\ \\\$
$E = I \times \pi {\left( {\dfrac{d}{2}} \right)^2} \\\ \\\$
d = 2 mm = $2 \times {10^{ - 3}}m$ [As 1 m = 1000 mm]
Substituting the values:

$E = \dfrac{{60}}{{4 \times \pi \times {{\left( 4 \right)}^2}}} \times \pi \times \dfrac{{{{\left( {2 \times {{10}^{ - 3}}} \right)}^2}}}{4}$ J/s
E = $9.375 \times {10^{ - 7}}$ J/s
Energy is also given by : $\dfrac{{hc}}{\lambda }$ [From photoelectric effect]
And for n number of particles: $\dfrac{{nhc}}{\lambda }$ where,
n = number of photons
h = Planck’s constant
$\lambda$ = wavelength
$E = \dfrac{{nhc}}{\lambda }$
Calculating for n:
$n = \dfrac{{E\lambda }}{{hc}}$
Substituting the values:
$n = \dfrac{{\;9.375 \times {{10}^{ - 7}} \times 600 \times {{10}^{ - 9}}}}{{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}$
n = $2.84 \times {10^{12}}$ photons/sec
Number of photons entered in 1 second = $2.84 \times {10^{12}}$
Number of photons entered in 0.1 seconds = $2.84 \times {10^{12}} \times 0.1$
=$2.84 \times {10^{12}} \times \dfrac{1}{{10}}$
= $2.84 \times {10^{11}}$

Therefore, the number of photons entering in the eyes for the given set up is $2.84 \times {10^{11}}$ photons, option B)

Note: The units are very important to remember and used appropriately:
Energy (E) = Joule / second (J/S)
Number of photons (n) = photons / second
Intensity (I) = Watt / ${(metre)^2}$ ($W/{m^2}$)