Im trying really understand how this is solved for my finals coming up, so if yo
ID: 2019442 • Letter: I
Question
Im trying really understand how this is solved for my finals coming up, so if you could please write out ALL of your steps with the final answer for each i would appreciate it very much!! Thanks in advance!!
Part 1: The de Broglie wavelength of a .060 Kg golf ball is 4.28 x 10-24m;
What is its speed? (h=6.63 x 10-34J.s)
a.) 15 m/s
b.) 26 m/s
c.) 31 m/s
d.) 48 m/s
Part 2: If the momentum of an electron is measured to be 3.2 x 10-23kg.m/s with an uncertainty of 0.5%, what is the minimum uncertainty in its position?
a.) 2.6 x 10-4 m
b.) 3.3 x 10 4 m
c.) 0.63 x 10-4m
d.) 1.1 x 106 m
e.) 7.6 x 10-4 m
Part 3: Find the longest wavelength (nm) of the photon emitted in the Paschen Series:
a.) 1282
b.) 1875
c.) 1923
d.) 2251
Part 4: The energy of the emitted photon in eV:
a.) 0.969
b.) 0.663
c.) 0.646
d.) 0.552
thank you so much in advance!! i know that it is a really long problem but i have exams this week and i want to be able to understand all of the practice test problems thoroughly. I really appreciate your help!!
Explanation / Answer
Part 1:
The deBroglie wavelength is
= h/p, so p = h/ and therefore
v = h/m = 6.63*10^-34 Js/(0.06 kg)(4.28*10^-24m) = 2.58*10^-9 m/s
Note: In order to get 26 m/s, answer b), you would need a deBroglie wavelength of 4.28*10^-34 m, not 10^-24m. Maybe you made a mistake writing down the numbers, or I made a mistake calculating the speed.
Part 2:
Use the Heisenberg principle of uncertainty. Unfortunately, there are two different uncertainty principles, and I dont know which one you use:
xp or xp /2, I'm gonna use the one that equals .
If the uncertainty of p is 0.5%, and the value of p is 3.2*10^-23 Ns, then
p = (0.005)(3.2*10^-23Ns) = 1.6*10^-25 Ns, so
x /p = (1.055*10^-34 Js)/(1.6*10^-25 Ns) = 6.59*10^-10 m
If you had used /2, you would have gotten 3.3*10^-10 m
Note: Once again, there is a decimal error in this. I don't understand why the numbers aren't matching up, but these are pretty simple calculations, so I would run through your numbers again.
Part 3:
The Paschen series starts at n = 3. The largest wavelength comes from the n = 4 -> n = 3 transition, so
E = 13.6 eV (1/(n1)^2 - 1/(n2)^2) = 13.6 eV (1/9 - 1/16) = 0.661 eV = hc/, so
= hc/E = (4.136 * 10^-15 eVs)(3*10^8 m/s)/(0.661 eV) = 1.877 * 10^-6 m = 1877 nm
The correct answer is b)
Part 4: As shown above, the energy of the emitted photon is 0.661 eV. The correct answer would be b).
I hope you figure out the discrepancies, sorry I couldn't work the numbers, but I would give the questions a good look over to make sure that the numbers you included were correct. I hope this helps either way. Good luck,.