Question

How many photons are contained in a burst of yellow light (589 nm) from a sodium lamp that contains 623 kJ of energy?

Answer 1

Answer:

1.847*10^24 photons

Explanation:

Generally, a photon has certain amount of energy which can be calculated using the equation below:

E = (h*c)/λ

Where:

E = energy of the a photon (J)

h = Planck's constant = 6.626*10^-34 (J.s)

c = speed of light = 299792458 (m/s)

λ = wavelength = 589 nm = 589*10^-9 m

Therefore:

E = (6.626*10^-34 * 299792458)/589*10^-9 = 3.373*10^-19 J

The energy of one photon is equivalent to 3.373*10^-19 J. Therefore, to find the number of photons in 623 kJ of energy, we have:

Number of photons = 623000/3.373*10^-19  = 1.847*10^24 photons

Thus, there are 1.847*10^24 photons in 623 kJ of energy.

Answer 2

First we need to find the energy of one photon with a wavelength of 589 nm. E = hc / wavelength E = (6.63 x 10^{-34} J s)(3 x 10^8 m/s) / (589 x 10^{-9} m) E = 3.3769 x 10^{-19} Joules To find N, the number of photons, we need to divide the total energy by the energy of each photon. N = 623000 J / 3.3769 x 10^{-19} Joules N = 1.84 x 10^{24} photons There are 1.84 x 10^{24} photons in the burst of yellow light.