Question

Question #baf4c

Answer 1

`30.8%`

Explanation:

The idea here is that you need to find the energy of a single photon associated with that given wavelength, then use the total energy emitted by the sample to figure out exactly how many photons were emitted.

Once you know that, use the molarity and volume of the solution to determine how many moles of naphthalene, `"C"_10"H"_8`, were present in the sample.

Finally, convert the moles of naphthalene to molecules and compare the number of photons with the number of molecules.

So, according to the Planck - Einstein relation, the energy of a photon is proportional to its frequency

`color(blue)(|bar(ul(color(white)(a/a)E = h * nucolor(white)(a/a)|)))`

Here

`E` - the energy of the photon
`nu` - its frequency
`h` - Planck's constant, equal to `6.626 * 10^(-34)"J s"`

As you know, frequency and wavelength have an inverse relationship described by the equation

`color(blue)(|bar(ul(color(white)(a/a)nu * lamda = c color(white)(a/a)|)))`

Here

`lamda` - the wavelength of the photon
`c` - the speed of light in a vacuum, usually given as `3 * 10^8"m s"^(-1)`

Use this equation to find a relationship between the energy of the photon and its wavelength

`lamda * nu = c implies nu = c/(lamda)`

Plug this into the P - E relation to get

`E = h * c/(lamda) = (h * c)/(lamda)`

Plug in your values to get the energy of a single photon of wavelength `"349 nm"` - do not forget to convert this from nanometers to meters!

`E = (6.626 * 10^(-34)"J"color(red)(cancel(color(black)("s"))) * 3 * 10^8color(red)(cancel(color(black)("m")))color(red)(cancel(color(black)("s"^(-1)))))/(349 * 10^(-9)color(red)(cancel(color(black)("m")))) = 5.696 * 10^(-19)"J"`

If this is the energy of a single photon of wavelength `"349 nm"`, it follows that the total energy of `"14.8 J"` corresponds to

`14.8 color(red)(cancel(color(black)("J"))) * "1 photon"/(5.696 * 10^(-19)color(red)(cancel(color(black)("J")))) = 2.598 * 10^(19)"photons"`

Now that you know how many photons were emitted, focus on finding how many molecules of naphthalene were present in solution.

You know that

`color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/V_"solution" implies n_"solute" = c * V_"solution")color(white)(a/a)|)))`

Plug in your values to get

`n_(C_10H_8) = "0.140 mol" color(red)(cancel(color(black)("L"^(-1)))) * overbrace(1.00 * 10^(-3)color(red)(cancel(color(black)("L"))))^(color(purple)("volume in liters"))`

`= 1.40 * 10^(-4)"moles C"_10"H"_8`

As you know, one mole of any substance contains `6.022 * 10^(23)` molecules of that substance.

`color(blue)(|bar(ul(color(white)(a/a)"1 mole" = 6.022 * 10^(23)"molecules"color(white)(a/a)|))) ->` Avogadro's number

Use Avogadro's number as a conversion factor to help you find how many molecules of naphthalene you have in that many moles

`1.40 * 10^(-4)color(red)(cancel(color(black)("moles C"_10"H"_8))) * overbrace((6.022 * 10^(23)"molec.")/(1color(red)(cancel(color(black)("mole C"_10"H"_8)))))^(color(purple)("Avogadro's number")) = 8.431 * 10^(19)"molec"`

The percentage of the naphthalene molecules that emitted a photon will thus be equal to

`"% emitted" = (2.598 * color(red)(cancel(color(black)(10^(19)))))/(8.431 * color(red)(cancel(color(black)(10^(19))))) xx 100 = color(green)(|bar(ul(color(white)(a/a)"30.8%"color(white)(a/a)|)))`

The answer is rounded to three sig figs.