Hi! Some help with this problem would be much appreciated.
The overlap integral is defined as ##S = \int \phi_A (\mathbf{r}_A) \phi_B (\mathbf{r}_B) \,d\mathbf{r}##. For the two orbitals, I have that
$$\phi_A = \frac{1}{\sqrt{\pi}} \Big( \frac{1}{a_0} \Big)^{3/2} e^{-r_A / a_0}$$
for the 1s...
Hi. I'm really stuck with this problem and would appreciate some help.
For example, if i take the total intensity from the ##^2\text{P}_{3/2}## level, i get ##a+b##. Since ##b## is 9 times larger than ##a##, i get that the total intensity is ##10a##. This should then be proportional to the...
Oh, I meant like using the approximation that ##3 \cdot 10^{50} \approx 10^{50}## as valid, while an approximation of ##3 \cdot 10^{-10} \approx 10^{-10}## would be invalid (in essence, for small numbers one shouldn't remove factors like that). The Taylor expansion have indeed been the most used...
I see! I thought about Taylor expanding after performing the integral, but actually expanding the integrand! Ok, Taylor expanding ##\left(\ f(x) = f(a) + f'(a)(x-a) + \frac{1}{2}f''(a)(x-a)^2 + ...\ , \text{with }a=0\right)## the integrand:
\begin{align*}
\frac{d}{dr} \left( r^2 e^{-2r/a_0}...
I can get the approximation ##1-e^{-2r_b/a_0}##, which seems okay-ish to me since it at least is 0 at ##r_b=0## and approaches 1 as ##r_b## approaches infinity, even though the approximation is only supposed to be used for small ##r_b##. I only used the reasoning that $$r_b^2 + a_0 r_b +...
Hi. I would love if someone could check my solution since me and the answer sheet I found online don't agree.
The probability is given by the triple integral
\begin{align*}
\int_0^{r_b} \int_0^{2\pi} \int_0^\pi |\psi (r)|^2 r^2 \sin{\theta} \,d\theta \,d\phi \,dr &= \frac{1}{\pi...
I think the problem has been WAY overanalysed. I think it's just a problem involving an initial string of length ##L## and frequency ##f##, that you subdivide into two parts and calculate their frequencies. At least, that would be my experience with homework problems.
For any deeper analytics...
From my understanding of the problem, ##f_2## is your frequency for the string of length ##2/3L##. Try to isolate it on one side of the equal sign and see what frequency you get.
Then do the problem for the string of length ##1/3L## and see what frequency you get for that.
To me, your equation for the volume ##V## looks fine, so if you don't get the correct solution I think you may have made a mistake later during your evaluation of the integrals. In particular, the integral ##\int_{1}^{e}ln^2y \ dy## looks a little nasty. Have you tried partial integration...
Alright, here goes. To recap: ##s = -\alpha t##, giving ##L = \frac{1}{2} m (\alpha^2 + \alpha^2 t^2 \dot{\theta}^2) - m g \alpha t \cos{\theta}##. Then the Hamiltonian becomes:
\begin{align*}
H =&\ \frac{\partial L}{\partial \dot{\theta}} \dot{\theta} - L\\
=&\ m \alpha^2 t^2 \dot{\theta}^2 -...
Ok, so using ##s = -\alpha t## to rewrite the Lagrangian as ##L = \frac{1}{2} m(\alpha^2 + \alpha^2 t^2 \dot{\theta}^2)## makes ##s## disappear from the equation and makes it only dependent on time and ##\dot{\theta}##. And then ##\theta## becomes the only coordinate to use in the equation for...
Hello!
I need some help with this problem. I've solved most of it, but I need some help with the Hamiltonian. I will run through the problem as I've solved it, but it's the Hamiltonian at the end that gives me trouble.
To find the Lagrangian, start by finding the x- and y-positions of the...