HW 1B.5

Moderators: Chem_Mod, Chem_Admin

005388369
Posts: 73
Joined: Sat Sep 28, 2019 12:16 am

HW 1B.5

Postby 005388369 » Tue Oct 15, 2019 9:58 am

The gamma ray photons emitted by nuclear decay of technetium-99 atom used in radoopharmaceuticals have an energy of 140.511 keV. Calculate the wavelength of these gamma rays. In the solutions manual, they convert keV to Joules and I understand why, but why did they multiply 140.511 by 10^3?

SarahSteffen_LEC4
Posts: 67
Joined: Sat Aug 24, 2019 12:17 am

Re: HW 1B.5

Postby SarahSteffen_LEC4 » Tue Oct 15, 2019 10:20 am

I believe they multiplied that number by 10^3 because of the fact that it is keV, thus to get from keV to just eV you would need to multiply that number by 10^3.

Jamie Hwang 2F
Posts: 104
Joined: Thu Jul 25, 2019 12:16 am

Re: HW 1B.5

Postby Jamie Hwang 2F » Tue Oct 15, 2019 6:30 pm

In the solutions manual, the answer for this question has 5 sig figs and I was confused on how they got 5 sig figs (the given value in the question has 6 sig figs, and the constants used in this equation (like Planck's, speed of light, conversion from ev to J) have either 3 or 4. I don't know if we will be tested so strictly on the sig figs but just in case can anyone explain this?

Anthony Hatashita 4H
Posts: 103
Joined: Wed Sep 18, 2019 12:21 am

Re: HW 1B.5

Postby Anthony Hatashita 4H » Thu Oct 17, 2019 6:17 pm

So do we have to convert keV to J for all problems or is it specific to this one?

Annie Chantasirivisal_4G
Posts: 114
Joined: Wed Sep 18, 2019 12:21 am

Re: HW 1B.5

Postby Annie Chantasirivisal_4G » Thu Oct 17, 2019 9:56 pm

Anthony Hatashita 4H wrote:So do we have to convert keV to J for all problems or is it specific to this one?


Since the problem is asking for the wavelength, you would use the formula (lambda)=ch/e out of combining e=hv and c=v(lambda).
The energy (140.511 keV) will be the denominator, but we'd want the units to cancel out from planck's constant in the numerator, which is why you should convert from keV to J.

Daniel Martinez 1k
Posts: 50
Joined: Wed Sep 18, 2019 12:16 am

Re: HW 1B.5

Postby Daniel Martinez 1k » Thu Oct 17, 2019 10:11 pm

Yes. You have to convert keV to J before starting the problem.

ramiro_romero
Posts: 90
Joined: Sat Sep 07, 2019 12:16 am

Re: HW 1B.5

Postby ramiro_romero » Fri Oct 18, 2019 12:57 am

I converted KeV directly to Joules so I also don't understand the ned to multiply KeV by 10^3. However, I am also very confused as to how these units cancel and my final answer was lambda=8.831pm.

I multiplied plancks constant by c and got 1.9878x10^-25. I then divided the product by energy in joules (2.251x10^-14) for lambda=8.831pm.

Micah3J
Posts: 100
Joined: Tue Oct 08, 2019 12:16 am

Re: HW 1B.5

Postby Micah3J » Fri Oct 18, 2019 1:45 am

ramiro_romero wrote:I converted KeV directly to Joules so I also don't understand the ned to multiply KeV by 10^3. However, I am also very confused as to how these units cancel and my final answer was lambda=8.831pm.

I multiplied plancks constant by c and got 1.9878x10^-25. I then divided the product by energy in joules (2.251x10^-14) for lambda=8.831pm.



You have to multiply by 10^3 because of the K in front of the eV. The K stands for Kilo. eV is just the abbreviation for electrical volts.

005321227
Posts: 90
Joined: Sat Sep 07, 2019 12:15 am

Re: HW 1B.5

Postby 005321227 » Fri Oct 18, 2019 11:37 am

you must multiply by 10^3 to convert to the correct units


Return to “Photoelectric Effect”

Who is online

Users browsing this forum: No registered users and 1 guest