Question

A radiation detector exposed to sunlight records the energy falling on a given area each unit time.f such a detector has a reading of 0.698 cal cm2 min1, how many photons of sunlight are striking each cm2 in one minute? Assume the average wavelength of sunlight is 460. nm. (4.184 J 1 calorie)A. 1.62 x 1018 photonsB. 4.32 x 1019 photonsC.6.76 x 1018 photons D 2.31 x 1018 photons E 9.63 x 1019 photons

Answer 1

Answer:

C. 6.76 x 10^18 photons

Explanation:

First, we need to find the energy falling on 1 cm² in 1 minute. It is given to be:

Energy = E = 0.698 cal

Converting it into Joules:

E = (0.698 Cal)(4.184 J/1 Cal)

E = 2.92 J

Now, for the number pf photons, we know that this energy from sun is given by the formula:

E = nhυ

where,

h = Plank's Constant = 6.625 x 10^-34 J.s

n =  number of photons

υ =  frequency of wave

But,

υ = c/λ

where,

c = speed of light = 3 x 10^8 m/s

λ = wavelength = 460 nm = 460 x 10^-9 m

Therefore,

E = nhc/λ

n = Eλ/hc

n = (2.92 J)(460 x 10^-9 m)/(6.625 x 10^-34 J.s)(3 x 10^8 m/s)

n = 6.76 x 10^18