The equation F= ke(|q1q2|)/r2 looks good.
So if I plug in 3.204 × 10^-19 coulombs for q1 and -3.204 × 10^-19 coulombs for q2 (because O2- has a net charge equal to -2 times the charge of an electron and Cu2+ has a net charge equal to twice the charge of an electron), then I get
2.307*10-8...
Well I don't know the force equation, my teacher only gave us the equation for bonding energy...
Perhaps since energy=force*distance we can find force by dividing our energy equation by some distance?
I'm still stuck but I see now that z_1= 2 and z_2=-2.
Any more help?
Homework Statement
Calculate the attractive force between a pair of Cu2+ and O2- ions in the ceramic CuO that has an interatomic separation of 200pm.
Homework Equations
E_A= -\frac{(z_1\cdot e)(z_2\cdot e)}{4\pi\cdot\epsilon_o\cdot r}
Where z_1 and z_2 are the valences of the two ion...
Homework Statement
Zinc has a density of 7.17 Mg/m^3. Calculate (a) the number of Zn atoms per cm^3, (b) the mass of a single Zn atom and (c) the atomic volume of Zn.
Homework Equations
atomic mass of zinc = 65.39 g/mol
The Attempt at a Solution
For part (a) I use the fact that...
Homework Statement
Find the energy of a He+ electron going form the n=4 state to the n=2 state.
Homework Equations
E_n=\frac{m\cdot e^4 \cdot z^2}{2n^2 \cdot \hbar^2}
Where m= mass of electron, z= atomic number, e= charge of an electron, n is the energy level.
^ I think those are...
Homework Statement
Using the Bohr model of the atom, compute the energy in eV of the one electron in Li2+.
Homework Equations
E_n=\frac{m\cdot e^4 \cdot z^2}{2n^2 \cdot \hbar^2}
Where m= mass of electron, z= atomic number, e= charge of an electron, n is the energy level.
^ I think...
"Period" is the time it takes for the particle to make one cycle. The particle traveling from 0 to A takes 1/4 of the time it would take to travel the whole period.
But I don't see how we can find the time it takes the particle to travel from 0 to A? We have this equation that gives us time...
If ##\frac{dx}{dt}=a+x## then ##\frac{1}{a+x}dx=dt## (whatever that means), so ##\int \frac{1}{a+x} dx = \int dt = t##. So ##ln(a+x)=t##.
If ##\frac{dx}{dt}=x^2## then ##\frac{1}{x^2}dx=dt## (once again idk what exactly i'm doing there), so ##\int \frac{1}{x^2}dx = \int dt##. So...
I'm afraid I don't get what it means to take ##\frac{dx}{dt}=f(x)## and turn it into ##\frac{1}{f(x)}dx=dt.
And thus I don't get what it means to then take the integral of both sides of that...
can you help me make sense of what those procedures mean?
##\frac{dx}{dt}=f(x)##
##\frac{1}{f(x)}dx=dt##
##\int f(x)^{-1} dx = \int 1 dt=t##
is that right?
That would give us time as a function of velocity which can give us velocity as a function of time?
Then we can integrate that with respect to t to find displacement as a function of time...
Since the input into our velocity equation we found is x not t, I should edit:
... to say:
See how as we make Δt smaller and smaller, the above becomes a better and better approximation of displacement x as a function of time? But I don't know how to turn that method into an integral...
Homework Statement
A particle oscillates with amplitude A in a one-dimensional potential that is symmetric about x=0. Meaning U(x)=U(-x)
First find velocity at displacement x in terms of U(A), U(x), and m.
Then show that the period is given by ##4\sqrt{\frac{m}{2U(A)}}\int_0^A...