Asked by Feather
An e−1 is released by an electron microscope at 3.00 × 106 m/s.
Determine in meters using the de Broglie relation
Substitute kgˑm2/s2 for J and show all units.
Then, convert to pm and show the conversion factor.
Determine the kinetic energy using EK = (1/2)mv2, where v is velocity.
Show all units.
So would it be 9.10938 x10^-31 j Gould would you convert that?
Determine in meters using the de Broglie relation
Substitute kgˑm2/s2 for J and show all units.
Then, convert to pm and show the conversion factor.
Determine the kinetic energy using EK = (1/2)mv2, where v is velocity.
Show all units.
So would it be 9.10938 x10^-31 j Gould would you convert that?
Answers
Answered by
DrBob222
I'm a little unsure of what you want. I assume that "box" I see and can't read is lamda. lambda = h/mv
So you plug in h, you have m right for the electron, and you have v listed, solve for wavelength.
If you want KE use KE = 1/2 mv^2 and units will be in J.
If you want to convert lambda in meters to pm, that is ?m x (10^12 pm/m) = ?
So you plug in h, you have m right for the electron, and you have v listed, solve for wavelength.
If you want KE use KE = 1/2 mv^2 and units will be in J.
If you want to convert lambda in meters to pm, that is ?m x (10^12 pm/m) = ?
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