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Numerade Educator



Problem 42 Medium Difficulty

In the Bohr model of the hydrogen atom, the electron moves in a circular orbit of radius $0.529 \times 10^{-10} \mathrm{m}$ around the proton. Assuming the orbital angular momentum of the electron is equal to $h / 2 \pi,$ calculate (a) the orbital speed of the electron, (b) the kinetic energy of the electron, and (c) the angular frequency of the electron's motion.


A. $2.19 \times 10^{6} \mathrm{m} / \mathrm{s}$
B. $2.18 \times 10^{-18} \mathrm{J}$
C. $4.13 \times 10^{16} \mathrm{rad} / \mathrm{s}$


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Video Transcript

in exercise. We have boards model for the hydrogen, Adam, on which the Electra, um, in red orbits the proton in blue. And we have that the radios off the orbit is this. 0, 00.5 29 time sent to the Manus 10 m. The mass of the latter room is this, and we can express the angular momentum off the Elektrim in this orbit as age over two. Pi on each age is plank's constant that I wrote here. So this exercise wants us to calculate what is the angular frequency off the electron's motion? What is the connect IQ? Energy off the electron. And finally what? It's the orbital speed off the electoral. So right here What we watch find we watch fight omega. We want to find do orbital velocity, and we want to find the connected energy off the election. Okay, so it's starting with, um Omega. Okay, we have that. Omega is just the angular momentum over the moment of the nation. This is from the definition of angular momentum, and in our case, the moment of inertia we treat the electron as a particles. At the moment of inertia is the moment of inertia of a particle. So the mass off the electoral times, the radios off the orbit squared, and we have that omega is equal to age over two pi times M e r squared and you find this to be equal to 4.13 time sin to 16 radiance per second. So this is the first answer. So now we want to find what is the orbit orbital velocity. So we have that there is a relation between the orbital velocity and the angular velocity. It is developed the orbital velocity. It is the angler velocity times, the orbital ray GIs. And, uh, this is just 4.13 taking the answer off the last exercise time send to the 16 times the radios, which is just 0.5 29 Time san to them. Understand? Those values are all given, and we have that The velocity is just on to 0.19 time stand to the 6 m per second and finally we want to calculate the kinetic energy. So the kind of take energy is just I Omega squared, okay, on which we were treating, uh, the election as a point particle. So we have that the kinetic energy is m e r squared times, omega squared. So once again, who submits to the values that we have our m e and omega that we calculated and we have that this is equal to 9.1 09 time sin to the minus 31 times R squared 0.5 29 times 10 to the minus Stan times are right here, here, in this bottom line Um, sorry. Squared R squared Omega squared, which is the answer off the first exercise. So four point 13 times sin should a 16 squared And we find a kinetic energy to be eagle to, um, 2.18 times. I'm sorry. This it is this over to so over to times have okay, so can extract energy is the moment of inertia times uh, angular velocity squared over two. And we find it to be, uh, 2.18 times 10 to the minus 18 ju. And this is the final answer off the exercise

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