Standing waves mean that if orbit is equal to a whole number of wavelengths, then a standing wave is formed. This means that electrons of those particular wavelengths can exist in the atom. Since different wavelengths correspond to different energies, only those energy states are possible for electron orbit, meaning that discrete energy levels exist.
Your answer is fine for VCE purposes, but I'm going to say again: this model is heavily oversimplified. The Bohr model of the atom only works for hydrogen atoms, where there is only one electron. Even for helium, the fact that you have two electrons and repulsions between electrons means you'll have problems. Indeed, Bohr was utterly unable to do anything about atoms with more than one electron.
hello,
can someone please explain this question?
why can't i use"
mgh = (1/2)kx^2
1 x 10 x 0.3 = (1/2) k (0.3)^2
thank you
Let me ask you this: what scenario does the question ask for? It's NOT an energy question because the masses are stationary at both times. This means the net force = 0. An extra extension of 30 cm counters the addition of 1 kg of mass. So k*0.3m = 1kg * g = 9.8 N
k = 9.8/0.3 N/m or around 33 N/m
I'll explain why you can't use energy. When you use mgh = 1/2 kx^2, you are saying the spring energy change equals the gravitational energy change. As mechanical energy is conserved and as the total potential energy seems to have remained constant, the kinetic energy must also have remained constant. Initially, before adding a mass, the spring wasn't moving. Therefore the kinetic energy initially is zero. Solving mgh = 1/2 kx^2 thus solves for the next position where the spring isn't moving either. This is all assuming NO EXTERNAL INFLUENCE! Conservation of energy only applies to a closed system.
Let me ask you this. If you got a spring, attached a mass to it and let go, what would happen? It'd oscillate up and down like a slinky. At its lowest point where the kinetic energy is momentarily zero, the net force isn't zero. It was moving down before, it's now not moving. Clearly the net force is going straight up. But in the diagram your mass is remaining stationary, so the net force is zero. So evidently, the stationary point you solve for with mgh = 1/2 kx^2 is a different point to the point where you can just hang the mass indefinitely.
The solution to this dilemma is: once you put the new mass on, you yourself have to then gradually pull the spring down until it stretches a bit more and the extra stretch of the spring balances the extra weight force. This is when you let go of the spring and nothing happens. See how you had to actively hold the spring? The system is no longer closed. Conservation of energy no longer applies.
In summary, use net force because the spring isn't moving and is staying still. That's not an energy question.