Textbook Problem 1B.5

[Nimisha_Seshadri_3C]

The γ-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 γ-rays.

[005801694]

For this problem, you want to use the equation E=(hc)/lambda to calculate lambda. First convert the given energy in keV to joules by multiplying 140.511 by 1000 to get it into eV then multiply that by 1.602*10^-19 J since 1eV=1.602*10^-19 J. Use that value of E and the constants h and c to solve for the wavelength.

[405745446]

Follow the steps as mentioned above ^ but I prefer doing the two equations separate instead of combining them (referring to E=(hc)/lambda) . In case you also prefer it as well,
After converting 140.511keV to eV and then to J:
Use E= hv where E is the number you get from converting to 140.511keV to J, h is planck's constant and v is frequency. Solve for v (frequency).

After getting frequency, plug it into
c= lambda (v)
where c is the speed of light and solve for lambda(wavelength)

[Erin Chin 1L]

Hi! So first, use the equation E=hc/lambda because you're given the energy of the photon in the problem and you want to solve for wavelength. You can rearrange this equation to hc/E(photon)=lambda. Since the given energy is in keV, you want to convert it to J. First you would multiply it by 1000 to get to eV. Then, multiply that value by 1.602*10^-19 J in order to get the energy in joules. Now you have all the values so you can plug them into the equation. Once you solve the equation, your value is in m so you need to convert it to pm because of γ-ray, so divide the value by 10^-12. Hope this helps!