- Which one do I need? The regular N router is less expensive than the later, but I want to buy the right one. What is benefit to the router with 1 gb?
Answer by Frabz
Model and Brand?
- Let N be an arbitrary integer. What is the least number of weights that can be used on
a scale pan to weigh any integral number of pounds from 1 to N inclusive, if the weights
can be placed in either of the scale pans?
Answer by Brian
The weights you need to do this are weights of powers of 3: 1,3,9,27,81,…
If you need to measure up to N, then you will need log_3(2*N+1) weights.
- I know the N shell can hold a max of 32 electrons, but what about when it is just a period 4 element? I think the outermost shell of period 4 elements is the N shell, so there must be some special rule.
Answer by sciteacher
Each level follows the rule e=2n^2
so level 1(n=1) e=2
level 2(n=2) e=2*4=8
n=3 so e=2(9)=18
n=4 so e = 2(16)
n=5 so e=2(25)=50 …………
- The original fermi level of the p and n-regions taken separately are different due to difference in concentration and type of doping. However, at the junction, the energy level diagram usually depicts a common fermi level while conduction and valency bands are displaced. How does that happen?
Answer by epidavros
Energy levels in materials are always relative. In other words, you can say that the conduction band is so many eV above the valence band, but it has no particularly absolute value.
The Fermi level is simply the energy level of the topmost occupied electronic state (at T = 0). Thid once again is relative – you can say where it is with respect to say the valence band.
Until, that is, you join the two materials. When you do this the energy of the topmost occupied state on the n side *must* be the same as that on the p side because the materials are in contact. If it were not, then electrons would flow into the lower energy states across the boundary until they were.
Clearly, for these two levels to be the same across the boundary, the energy bands measured relative to the Fermi level must change, because the band structure must be continuous across the boundary. For this to happen the conduction and valence bands must curve. The origin of this curvature is, of course, and electical potential energy gradient across the junction. This is what gives rise to the intrinsic junction bias.
- What is the energy difference in Joules between the n=6 state and the n=2 state in a hydrogen atom?
Please help me! I just want to know how to solve this problem? Thank you so much!
Answer by Anonymous
You have to solve certain equations.
1) mv^2/r = (4*pi*epsilon)^(-1)*q^2/r^2, i.e. centripetal force is equal to coulomb’s force. Of course, we’re assuming circular orbit. Otherwise, both forces cannot be written as such. It is still a good approximation.
2) Note that an electron in orbit does not emit electromagnetic radiation (despite centripetal acceleration). A quantum approach is needed. The explanation is that angular momentum is quantised. So use L = r x p = mvr = n(h/(2pi)), since r is orthogonal to p and where h is Planck’s constant.
From these two facts you can discover that the energy of an electron in a hydrogen atom is equal to:
-13.6* (1/n^2) eV. It is negative because work must be done by the electron in order to escape the pull of the proton (gravitational analogy works here).
So the difference in energy is: -13.6/6^2 – (-13.6/2^2) eV = 13.6*(8/36) eV = 3.02 eV.
(Note that 1 eV = 1.6 * 10^-19 J, so 3.02 eV = 4.83 * 10^-19 J)