## Wavelength and Energy

$c=\lambda v$

Posts: 17
Joined: Wed Sep 21, 2016 2:59 pm

### Wavelength and Energy

Does anyone know how to solve problem numbers 23 and 25 on the chapter one homework? I've tried many times and I can't seem to figure them out!
Thanks!

Melanie_Wong_1K
Posts: 22
Joined: Fri Jul 22, 2016 3:00 am

### Re: Wavelength and Energy

For 1.23 remember that wavelength=speed of light constant/frequency, and frequency = Energy/Planck's constant, so if Y is wavelength (because I can't figure out how to type lambda) Y=c/v and v=E/h. So you just have to set it up as

Y=hc/E

and subsitute 6.626x10^(-34) for h, 3.00x10^(8) for c, and plug in 140.511 keV for E.

Christopher Reed 1H
Posts: 33
Joined: Wed Sep 21, 2016 2:57 pm
Been upvoted: 1 time

### Re: Wavelength and Energy

Hi!

Question 23 is asking you to calculate the wave length of these gamma waves given the energy in KeV (Kiloelectronvolts).

The first step is to convert from KeV to Joules, the SI unit for energy.

$140.511 KeV * \frac{1.6022x10^{-19} J}{1 KeV} = 2.513x10^{-14} J$

Now we have the energy of the gamma rays. For the following steps let c = speed of light, E = energy of the photon, h = planks constant, $\lambda$ = wavelength and $\nu$ = frequency.
We can calculate the wavelength from this point by recalling these two formulas:

$E=h\nu$
$c=\nu \lambda$

Through some rearranging and an eventual substitution we find that the wavelength can be calculated through the following equation:

$\lambda = \frac{hc}{E}$

Since we have already calculated the only non-constant (frequency) we can substitute our values in and solve.

$\lambda = \frac{(6.626x10^{-34}J*S)*(2.99792x10^{8}m*S^{-1})}{2.2513x10^{-14}J}$

All units cancel except meters and keeping in mind the 5 significant figures...

$\lambda = 8.8237x10^{-12}m$
--------------------------------------

For question 25 I think this formula will be of use to you:

$E = \frac{hc}{\lambda }$

For part A since we are already given the wavelength of the of the light we can plug straight into the above equation after converting nm to m.

$E = \frac{(6.62608x10^{-34}J*S)*(2.99792x10^{8}m*S^{-1})}{589x10^{-9}m}$

All units cancel except Joules and keeping in mind the 3 significant figures you are left with

$E = 3.37x10^{-19}J$

Keep in mind that this is the amount of energy emitted PER ATOM OF Na.

For part B you need to calculate how many atoms of Na you have. You do so using the molar mass of Na, which is 22.99g*mol-1, and recall Avogadro's number, 6.022x1023. In the following calculation I am converting grams to moles, then moles to atoms, and then atoms to Joules using the above calculation that told us one atom emits 3.37x10-19J.

$E = \frac{5.00 x 10^{-3}g Na}{22.99g*mol^{-1}} * \frac{6.022x10^{23}atoms}{mol}*\frac{3.37x10^{-19}J}{atom}$

All units cancel except Joules and keeping in mind 3 significant figures...

$E = 44.1 J$

For part C you are already given the amount of moles so you just need avogadro's number to find the number of atoms.

$E = \frac{6.022x10^{23}atoms}{mols}*\frac{3.37x10^{-19}J}{atom}$

This leaves you with joules per mol which is ok because the problem is asking how much energy ONE mole of Na atoms emits.

$E = 2.03x10^{5}J$

Sorry about the long post, but I hope I helped! If you have any questions about the above work please ask! (:

danae_blodgett_1H
Posts: 18
Joined: Sat Jul 23, 2016 3:00 am

### Re: Wavelength and Energy

Thank you so much, this problem was giving me trouble as well. Now I understand!