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

Answer 1

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

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#
#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#
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

Use the formula (E = \frac{hc}{\lambda}) to find the energy of one photon. Then, apply (P = nE) to determine the number of photons per second, where (P) is power (in watts) and (n) is the number of photons.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7