Question

At noon on a clear day, sunlight reaches the earth's surface at Madison, Wisconsin, with an average intensity of approximately 1.00 kJ·s−1·m−2. If the sunlight consists of photons with an average wavelength of 510.0 nm, how many photons strike a 8.90 cm2 area per second?

Answer 1

The number of photon ejected per second by the incident light is 2.28 x 10¹⁸ photons / second.

How to calculate the number of photons ejected?

The power of the incident light is calculated by applying the following formula as follows;

P = IA

where;

• I is the intensity of the light
• A is the area of the surface

The given parameters include

I = 1 kJ / m²s

A = 8.9 cm² = 8.9 x 10⁻⁴ m²

P = 1,000 J/m²s x 8.9 x 10⁻⁴ m²

P = 0.89 J/s

The energy of a single photon is calculated ad follows;

E = hf

E = hc/λ

where;

• h is Planck's constant
• c is the speed of light
• λ is the wavelength

E = (6.626 x 10⁻³⁴ x 3 x 10⁸ ) / (510 x 10⁻⁹)

E = 3.9 x 10⁻¹⁹ J/photon

The number of photon ejected per second by the incident light is calculated as follows;

n = P / E

n = (0.89 J / s) / ( 3.9 x 10⁻¹⁹ J/photon)

n = 2.28 x 10¹⁸ photons / second

Answer 2

2.282 ×
`10^(18)` photons strike

Step-by-step explanation:

Given data

average intensity = 1.00 kJ·s−1·m−2

wavelength = 510.0 nm

area = 8.90 cm²

to find out

how many photons strike

solution

we know energy of 1 photon i.e

= hc/ wavelength = 6.626 ×
`10^(-34)` × 3
`10^(8)` / 510
`10^(-9)`

energy of 1 photon = 3.9 ×
`10^(-19)` J

so now we know no of joules hitting is 8.90 cm²/sec

i.e = 1.00 kJ × 8.90 cm²

= 1000 × 8.90
`10^(-4)`

= 0.8900 J

so number of photon is

N = 0.8900 / 3.9 ×
`10^(-19)`

N = 2.282 ×
`10^(18)`

2.282 ×
`10^(18)` photons strike