## Homework Problem 1.23 [ENDORSED]

$c=\lambda v$

Dan_Jin_1K
Posts: 15
Joined: Wed Sep 21, 2016 2:59 pm

### Homework Problem 1.23

Can anybody walk me through how I can answer this question?

I'm confused as to where I should start.

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 these y-rays.

Makenna Vulgas 1G
Posts: 20
Joined: Wed Sep 21, 2016 3:00 pm

### Re: Homework Problem 1.23  [ENDORSED]

Hi! Here's a step by step of how I did the problem.

1. Convert 140.511 keV into eV, so multiply 140.511 by 1000.
= 140511 eV or 140.511x10^3 eV

2. Convert eV to J. (I had to look up the conversion because I didn't know it) 1eV = 1.6022x 10^-19 J
(140511 eV)x(1.6022x10^-19 J) = 2.2513x10^-14 J

3. Find the wavelength since we know E =2.2513x10^-14 J from the previous problem.
I combined E = hv and c= (lambda) v in order to solve for lambda.
v = E/h & lambda = c/v therefore lambda = ch/E
(3.0x10^8 m/s)(6.626x10^-34 Js)/ (2.2513x10^-14 J) = A wavelength of 8.8296x10^-12 m or 8.8296 pm

Hope that helped!

DBaquero
Posts: 13
Joined: Wed Sep 21, 2016 2:59 pm
Been upvoted: 1 time

### Re: Homework Problem 1.23

So the first thing you want to do is convert the 140.511 keV to Joules. There are 1.6022x10^-19 Joules in 1 eV so there are about 2.251x10^-14 J in 140.511 keV
(140.511 keV * 1000 eV/keV* 1.6022x10^-19 J/eV = 2.251x10^-14 J)
Now we know that we can calculate the frequency with Planck's constant. (2.251x10^-14 J / 6.626x10^-34 Js = 3.398x10^19 Hz)
Finally with the frequency all that's left is to calculate the wavelength using c (2.998x10^8 m/s / 3.398x10^19 Hz = 8.8237x10^-12 m)
We can convert that to 8.237 pm.
Hope that helped.