1B. 5

Moderators: Chem_Mod, Chem_Admin

Abigail Menchaca_1H
Posts: 104
Joined: Sat Sep 07, 2019 12:19 am

1B. 5

Postby Abigail Menchaca_1H » Sun Oct 20, 2019 6:16 pm

"The y-ray photons emitted by the nuclear decay of a technetium-99 atom used in radiopharmaceuticals have an energy of 140.511 keV. Calculate the wavelength of the these y-rays."
I'm not sure where to start with this problem.

Sydney Jacobs 1C
Posts: 59
Joined: Thu Jul 25, 2019 12:15 am
Been upvoted: 1 time

Re: 1B. 5

Postby Sydney Jacobs 1C » Sun Oct 20, 2019 6:28 pm

First, it is necessary to convert 140.511 keV into Joules. Since there are 1.6022 x 10^-19 J per eV, multiply (140.511 x 10^3 eV) x ( 1.6022 x 10^-19 J/eV) to obtain 2.2513 x 10^-14 J. Combining the formulas E=hv and c= v x lambda, solve for lambda= hc/E.
Using Planck's constant, the speed of light, and the energy in Joules, solve for wavelength= 8.8237 pm.

Kendall 3H
Posts: 52
Joined: Wed Sep 18, 2019 12:18 am

Re: 1B. 5

Postby Kendall 3H » Sun Oct 20, 2019 8:17 pm

The first step is to convert the energy units from keV to Joules.
- 140x10^3 eV x (1.6022 x10^19 joules/ 1 eV) = 2.2513 x 10^-14 Joules

Next you can use the formula wavelength= hc/E to find the wavelength
- wavelength= (6.626x10^-34 X 2.9x10^8)/ (2.2513x10^-14) which equals 8.82 x 10^-12m or 8.82pm

Return to “Properties of Light”

Who is online

Users browsing this forum: No registered users and 1 guest