Like

Report

(a) Use de Broglie's hypothesis to determine the speed of the electron in a hydrogen atom when in the $n=1$ orbit. The radius of the orbit is $0.53 \times 10^{-10} \mathrm{m} .$ (b) Determine the electron's de Broglie wavelength. (c) Confirm that the circumference of the orbit equals one de Broglie wavelength.

A. $v_{e}=2.2 \times 10^{6} \mathrm{m} / \mathrm{s}$

B. $\lambda=3.3 \times 10^{-10} \mathrm{m}$

C. $3.3 \times 10^{-10} \mathrm{m}$

You must be signed in to discuss.

Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

Simon Fraser University

Hope College

using the Burl J wavelength hypothesis and in equals one and the radius of orbit r equals 0.53 times 10 to the minus 10 meters for part A were asked to figure out the speed of the electron. Okay, so to figure out the speed of the electron, we're gonna use momentum. So momentum classically l is equal to the mass times the velocity times the radius r Assuming that this thing is orbiting in a circular pattern, this is also equally equal to from the liberal J hypothesis in a number of integers of the orbit times H plank's constant divided by two pi. So we can rearrange this equation to solve for the velocity V we find it's equal to in times H divided by two times pi times the radius Are you using an equals one amusing H equal to 6.63 times 10 to the minus 34 jewels Second in the value for the radius that were given. We find that the velocity is equal to approximately 2.2 times 10 to the six meters per second. So it's going pretty fast. Not quite relativistic speeds, though, since ah the speed of light is three times 28 meters per second, so it's still a few orders of magnitude smaller than the speed of light. So this is part a part B would ask us to find the wavelength now that we know to speed. So the wavelength Year, according to the do Parole J hypothesis, is equal to Planck's constant H, divided by the BoE mentum of the article or the mass times. The velocity of the particle, the mass here, since it's an electron would just be the mass of the electron 9.1 times 10 to minus 31. Playing these values in, we find that this is equal to 3.3 times 10 to the minus 10 meters could also call this, uh, I'm sending minus 10. Sorry about that. You could also call this 33 nanometers, but I'll just leave it in units of meters brass to confirm that the circumference of orbit is equal to approximately one labelling so classically If we assume that this thing is, uh, circular, the circumference C is equal to two pi r. So we can use the value for our that we were given plugging this value and we find that this is approximately equal to 3.3 times 10 to the minus 10 meters. So we confirmed that one wavelength is equal to the circumference. Linkenbach sending is the solution for part C.

University of Kansas

Atomic Physics