Normal 0 false false false EN-US JA X-NONE /* Style Definitions */table.MsoNormalTable{mso-style-name:"Table Normal";mso-tstyle-rowband-size:0;mso-tstyle-colband-size:0;mso-style-noshow:yes;mso-style-priority:99;mso-style-parent:"";mso-padding-alt:0in 5.4pt 0in 5.4pt;mso-para-margin:0in;mso-para-margin-bottom:.0001pt;mso-pagination:widow-orphan;font-size:10.0pt;font-family:Cambria;} A low-pressure mercury-vapor lamp has a characteristic emission line at 253 nm. Knowing that this lamp is putting out 11.8 watts of light energy, how many mercury atoms are emitted per second during operation?

So here we're told that a low pressure mercury vapor lamp has a characteristic emission line at 253 nanometers. Knowing that this lamp is putting out 11.8 watts of light energy, how many mercury atoms are emitted per second during operation? Alright. So, we're told that it gives us an emission line at 253 nanometers. Since we're dealing with nanometers, that must mean that this represents our wavelength. We're then told that we're dealing with watts. Remember, a watt is equal to joules per second. 11.8 watt is 11.8 joules per second. Joules are a form of energy and we have wavelength. Remember, they're connected together by the equation that energy of a photon is equal to Planck's constant times speed of light divided by wavelength. What we're gonna do now is we're going to plug in our wavelength into this equation to figure out the energy of a single photon. So our Planck's constant, we'll plug that in. Plug in speed of light. And now we need our wavelength to be in meters, not nanometers. So we have 253 nanometers. For every 1 nanometer, it's 10 to the negative 9 meters, so that's 2.53 times 10 to the negative 7 meters. So here, meters cancel out, seconds cancel out and we'll have joules per photon. So our joules per photon comes out to 7.86 times 10 to the negative 19 joules per photon. Now that we have joules per photon and we have joules per second, we can use that to isolate our mercury atoms per second. Remember here, we can substitute the term atoms and photons. So here we can say that this just represents joules per atom. Alright. So we want seconds per second. So we have 11.8 joules per one second times we're gonna put 7.86 times 10 to the negative 19 joules on the bottom, so the units can cancel out. And we'll put 1 mercury atom on top. So at the end, our ants will be in atoms per second which is what we're looking for. So that gives me 1.50 times 10 to the 19 mercury atoms per second. So that will be our final answer. So remember, for a question like this, just look at what the variables are and think back on what equations connect them together. Here, they're giving us wavelength. A watt deals with energy, so we're dealing with joules. Wavelength and our energy will be connected by this formula here. Starting out with that helps us to determine that we'll have joules per per atom, and then look and see what units they want you to isolate at the end, which is atoms per second, guide you to what cancels out at the end to get your final answer.

17. Fundamentals of Spectrophotometry

Properties of Light

Analytical Chemistry

17. Fundamentals of Spectrophotometry

Properties of Light

Struggling with Analytical Chemistry?

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Multiple Choice

A low-pressure mercury-vapor lamp has a characteristic emission line at 253 nm. Knowing that this lamp is putting out 11.8 watts of light energy, how many mercury atoms are emitted per second during operation?

7.86x10^-19 Hg atoms/s

4.73x10⁵ Hg atoms/s

1.50x10¹⁹ Hg atoms/s

1.50x10²⁸ Hg atoms/s

9.27x10^-18 Hg atoms/s

Verified step by step guidance

First, understand that the energy of a photon can be calculated using the equation: E = \frac{hc}{\lambda}, where E is the energy of the photon, h is Planck's constant (6.626 x 10^{-34} J·s), c is the speed of light (3.00 x 10^8 m/s), and \lambda is the wavelength of the light (253 nm or 253 x 10^{-9} m).

Convert the wavelength from nanometers to meters to ensure consistency in units. This is crucial for accurate calculations.

Calculate the energy of a single photon using the formula E = \frac{hc}{\lambda}. Substitute the values for h, c, and \lambda to find the energy in joules.

Next, determine the number of photons emitted per second by dividing the total power output of the lamp (11.8 watts, which is equivalent to 11.8 joules per second) by the energy of a single photon calculated in the previous step.

The result from the division will give you the number of photons emitted per second, which corresponds to the number of mercury atoms emitted per second, as each photon corresponds to one emission event from a mercury atom.

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