How Does Electron Kinetic Energy Compare to Hydrogen's Ground State?

  • Thread starter Thread starter Woopy
  • Start date Start date
  • Tags Tags
    Wavelength
AI Thread Summary
The discussion focuses on calculating the kinetic energy of an electron with a de Broglie wavelength equal to the diameter of a hydrogen atom in its ground state. The initial calculation yielded a kinetic energy of 1.17e-20 J or 0.0732 eV, which was incorrect. A key error identified was the use of the proton mass instead of the electron mass, which is significantly lighter. This mistake led to a substantial discrepancy in the results. Participants emphasized the importance of verifying fundamental constants in such calculations.
Woopy
Messages
149
Reaction score
0

Homework Statement



An electron has a de Broglie wavelength equal to the diameter of a hydrogen atom in its ground state.
What is the kinetic energy of the electron?

How does this energy compare with the ground-state energy of the hydrogen atom?

Homework Equations



λ = h /mv
diameter = 2 x bohr radius = 1.06e-10 m

The Attempt at a Solution



1.06e-10 = 6.63e-34/1.67e-27v
v = 3.75e3 m/s
.5(1.67e-27)(3.75e3)^2 = 1.17e-20 J = .0732 eV
Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully.
 
Physics news on Phys.org
Woopy said:

Homework Statement



An electron has a de Broglie wavelength equal to the diameter of a hydrogen atom in its ground state.
What is the kinetic energy of the electron?

How does this energy compare with the ground-state energy of the hydrogen atom?

Homework Equations



λ = h /mv
diameter = 2 x bohr radius = 1.06e-10 m

The Attempt at a Solution



1.06e-10 = 6.63e-34/1.67e-27v
v = 3.75e3 m/s
.5(1.67e-27)(3.75e3)^2 = 1.17e-20 J = .0732 eV
Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully.


Well, your electron mass doesn't look right at all. I suspect that you used the value for the proton mass rather than the electron mass. They're different by a factor of more than 1000.
 
hah that was it. Thanks
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top