Let's first write down our givens for this problem
Given
color(green)("Power" = 600 W or 600 J/s)Power=600Wor600Js
color(green)(lambda = 331.3" nm" or (3.313*10^-7" m"))λ=331.3 nmor(3.313⋅10−7 m)
Now let's try to establish a few things.
The mercury lamp is emitting a certain amount of "Energy"Energy per "second"second, defined as its "Power"Power. The source is coming from a monochromatic radiation with a wavelength of 3.313*10^-7" m"3.313⋅10−7 m. We are asked to find how many photons are being emitted from the lamp that is providing 600" J"600 J of energy per second.
---------------------−−−−−−−−−−−−−−−−−−−−−
color(blue)"Step 1: Figure out the energy associated with the photon"Step 1: Figure out the energy associated with the photon
We use the following formula
color(white)(aaaaaaaaaaaaaaa)color(magenta)(E = h*f)aaaaaaaaaaaaaaaE=h⋅f
Where
"E = Energy of the photon (J)"E = Energy of the photon (J)
"h = Planck's constant" (6.62*10^-34" J*s")h = Planck's constant(6.62⋅10−34 J*s)
"f = frequency of the photon" (1/s)f = frequency of the photon(1s)
But we aren't given the frequency; we are give the wavelength.
Well, we know that the speed of light is constant and given as 3.00*10^8m/s3.00⋅108ms and can be calculated by using the following:
color(white)(aaaaaaaaaaaaa)color(magenta)(c = f*lambda)aaaaaaaaaaaaac=f⋅λ
Where
c = "speed of light"(3.00*10^8m/s)c=speed of light(3.00⋅108ms)
f = "frequency of radiation" (1/s)f=frequency of radiation(1s)
lambda = "wavelength" (m)λ=wavelength(m)
Knowing this, we can isolate to solve for the frequency
color(white)(aaaaaaaaaaaaa)color(magenta)(c = f*lambda)->c/lambda = faaaaaaaaaaaaac=f⋅λ→cλ=f
We can replace "f"f in the energy of the photon equation with c/lambdacλ and solve.
color(magenta)(E = h*f)->E = (hc)/lambdaE=h⋅f→E=hcλ
- E =[(6.62*10^-34 J*cancel"s")(3.00*10^8cancel"m"/cancel"s")]/(3.313*10^-7 cancel"m")->color(orange)(5.99 * 10^-19" J")
color(blue)"Step 2: Use dimensional analysis to figure out photons emitted per s"
Well, what does this tell us? This tells us that 1 photon has 5.99*10^-19" J"
color(white)(aaaaaaaaaaaaaa)(1" photon")/(5.99*10^-19" J")
We were also told that the lamp produced 600" J" per second
color(white)(aaaaaaaaaaaaaaaaa)(600" J")/s
We can use dimensional analysis to figure out the number of photons emitted per second.
color(white)(aaaaa)(600 cancel"J")/s*(1" photon")/(5.99*10^-19 cancel"J")->color(orange)[(1.00*10^21" photons")/s]
color(orange)["Answer": (1.00*10^21" photons")/s
