Recent content by Linus Pauling

Nevermind got em

Really lost here... I just tried solving it using a problem in my book as an analogy. For the probability at x = 0: I said that the equation describing one half of the "big triangle" is 0.50(1-x/1nm), which is psi squared. Solving with x = zero then multiplying by two to account for the...

I know this is probably simple but I just don't see it. I don't see how to get an equation that's a function of x that I can plug my x values into. For example, when it asks me for the probability at x=o over a length L, I know that the integral of psi squared is just (1/2)bh = 0.5, but it can't...

I know that P will be psi(x) squared multiplied about the length L given in the problem. However, from the graph I can get the integral of psi squared, which is simply the area under the curve. How do I compute psi squared itself?

1. What is U(x)psin(x) in the interval 0 < x < L? (less than or equal to) 2. In previous problems I had found the following: Second deriv. of psi(x) = -(n*pi/L)2*C*sin(n*pi*x/L) U(x)*psin(x) = 0 3. From the above and the Shrodinger equation, I added U(x) = 0 to the second...

1. The figure shows the probability density for an electron that has passed through an experimental apparatus. What is the probability that the electron will land in a 2.40×10−2-mm-wide strip at: I'm then asked the probability of finding a particle at various spots on the x-axis. We'll go...

I am also asked (and am confused about) computing the probablity of finding thte particle between -1 and 1nm. Do I simply take the integral/area of the "bigger rectangle" that goes from -1 to 1nm, whose are is 2c^2, then divide by the total area 2.5c^2, i.e. 80% probability?

Ok, so the shape of the squared wavefunction is basically two smaller rectangles with height 0.25c^2 and base of length 1, and a bigger one with base of 2 and height c^2. Total area = the integral = 1 = (5/2)c^2, so c = 0.632 Correct?

I was asked for the energy and orbital angular momentum of a hydrogen atom in the 6f state. Thus, with n = 6 I found E to be -0.378, a straightforward calculation. I am confused about L, though. L = sqrt[l(l+1)]. Since n = 6, I thought l = n-1, so L would be sqrt(30)*h-bar, but the answer is...

Why h-bar? The uncertainty principle is delta(x)delta(p) greater than/equal to h/2

I'm lost. What equation describes teh shape of the wavefunction given in the picture? I Understand that I will square that function, then integrate between the boundaries setting the integral equal to one, then solving for c...

1. What is c? 2. 1 = integral of wavefunction2 from -infinity to +infinity 3. From the graph, the area above the x-axis is 2/3 the total area. I solved the following integral (int) from -1 to +1: 2/3 = int(c2dx) Obtaining 2/3 = 2c2 So c = sqrt(1/3) = 0.58 nm-1/2

1. A proton is confined within an atomic nucleus of diameter 3.80 fm. Estimate the smallest range of speeds you might find for a proton in the nucleus. 2. Uncertainty Principle 3. 3.80fm * mv = h/2 Soliving for h, using mass of proton = 1.67*10-27kg, I obtain v = 5.22*107. Then...

1. Magnetic resonance imaging needs a magnetic field strength of 1.5 T. The solenoid is 1.8 m long and 75 cm in diameter. It is tightly wound with a single layer of 2.20-mm-diameter superconducting wire. 2. B = mu0BI / l 3. Is there a way to do this without finding N? Either way...

1. The predictions of Rutherford's scattering formula failed to correspond with experimental data when the energy of the incoming alpha particles exceeded 32MeV. This can be explained by the fact that the predictions of the formula apply when the only force involved is the electromagnetic force...