Answer: a) for 150 Angstroms 6.63 *10^-3 eV; b) for 5 Angstroms 6.02 eV
Explanation: To solve this problem we have to use the relationship given by De Broglie as:
λ =p/h where p is the momentum and h the Planck constant
if we consider the energy given by acceleration tube for the electrons given by: E: e ΔV so is equal to kinetic energy of electrons p^2/2m
Finally we have:
eΔV=p^2/2m= h^2/(2*m*λ^2)
replacing we obtained the above values.
Answer:
The energy of an electron in an isolated atom depends on b. n only.
Explanation:
The quantum number n, known as the principal quantum number represents the relative overall energy of each orbital.
The sets of orbitals with the same n value are often referred to as an electron shell, in an isolated atom all electrons in a subshell have exactly the same level of energy.
The principal quantum number comes from the solution of the Schrödinger wave equation, which describes energy in eigenstates
, and for the case of an hydrogen atom we have:

Thus for each value of n we can describe the orbital and the energy corresponding to each electron on such orbital.
Answer:
Given that
D= 4 mm
K = 160 W/m-K
h=h = 220 W/m²-K
ηf = 0.65
We know that

For circular fin


m = 37.08


By solving above equation we get
L= 36.18 mm
The effectiveness for circular fin given as


ε = 23.52
That they travel in a vacuum. All other waves require a medium in which they wave.
Answer:
a = v^2/r centripetal acceleration
v = (a * r)^1/2 = (42 * .23)^1/2 = 3.1 m/s