A 600W mercury lamp emits monochromatic radiation of wavelength 331.3 nm. How many photons are emitted from the lamp per second?

1 Answer
May 10, 2017

Answer:

#(1.00*10^21" photons")/s#

Explanation:

Let's first write down our givens for this problem

Given
#color(green)("Power" = 600 W or 600 J/s)#
#color(green)(lambda = 331.3" nm" or (3.313*10^-7" m"))#

Now let's try to establish a few things.

The mercury lamp is emitting a certain amount of #"Energy"# per #"second"#, defined as its #"Power"#. The source is coming from a monochromatic radiation with a wavelength of #3.313*10^-7" m"#. We are asked to find how many photons are being emitted from the lamp that is providing #600" J"# of energy per second.

#---------------------#

#color(blue)"Step 1: Figure out the energy associated with the photon"#

We use the following formula

#color(white)(aaaaaaaaaaaaaaa)color(magenta)(E = h*f)#

Where
#"E = Energy of the photon (J)"#
#"h = Planck's constant" (6.62*10^-34" J*s")#
#"f = frequency of the photon" (1/s)#

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/s# and can be calculated by using the following:

#color(white)(aaaaaaaaaaaaa)color(magenta)(c = f*lambda)#

Where
#c = "speed of light"(3.00*10^8m/s)#
#f = "frequency of radiation" (1/s)#
#lambda = "wavelength" (m)#

Knowing this, we can isolate to solve for the frequency

#color(white)(aaaaaaaaaaaaa)color(magenta)(c = f*lambda)->c/lambda = f#

We can replace #"f"# in the energy of the photon equation with #c/lambda# and solve.

#color(magenta)(E = h*f)->E = (hc)/lambda#

  • #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#