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About LinkBacksP(r) = [rR(r)]^2,
R(r) =, a0 = 0.529 * 10^-10
Find P'(r).
This is an equation from quantum mechanics, I have tried putting log on both sides and pull the powers down, but I'm not sure whether I'm doing the correct thing, please help me. Thanks!!
P(r) = [rR(r)]^2,
R(r) =
Find P'(r).
This is an equation from quantum mechanics, I have tried putting log on both sides and pull the powers down, but I'm not sure whether I'm doing the correct thing, please help me. Thanks!!
Is it R(r), or P(r)? let's say P(r), and i'll call a0 aOriginally Posted by Sasuke12
P(r) = [rR(r)]^2,
R(r) =
Find P'(r).
This is an equation from quantum mechanics, I have tried putting log on both sides and pull the powers down, but I'm not sure whether I'm doing the correct thing, please help me. Thanks!!
P = (2 - r/a)[(e^-r/2a)/2a^(3/2)] .............let's try the product rule
=> P' = (-1/a)[(e^{-r/2a})/2a^(3/2)] + (2 - r/a)[(-1/2a)(1/2a)^(3/2) * e^{-r/2a}]
=> P' = -e^{-r/2a}/[a(2a)^(3/2)] - (2 - r/a)(1/2a)^(5/2)*e^{-r/2a}
and this can be simplified further
now let's do it by the way you suggested, by logarithmic differentiation (which i think would be more complicated than it has to be)
P = (2 - r/a)[(e^-r/2a)/2a^(3/2)]
=> lnP = ln(2 - r/a)[(e^-r/2a)/2a^(3/2)]
=> lnP = ln(2 - r/a) + ln[(e^-r/2a) - ln[2a^(3/2)]
=> lnP = ln(2 - r/a) - r/2a - ln[2a^(3/2)] ..............now differentiate (implicitly)
=> P'/P = 1/(2 - r/a)*(-1/a) - 1/2a
=> P'/P = -1/(2a - r) - 1/2a
=> P' = P(-1/(2a - r) - 1/2a)
=> P' = {(2 - r/a)[(e^-r/2a)/2a^(3/2)]}(-1/(2a - r) - 1/2a)
hey that's not so bad after all, at least in comparison to the first answer we got. of course, both methods can be simplified further, which you might want to do if you're going to do a lot more work with P'. Good luck
Nope. He got it right. This is an n = 2, l = 0 Hydrogen electron wave function. (Or at least the radial part of an n = 2 wavefunction.)P(r) = [rR(r)]^2,
R(r) =
Find P'(r).
This is an equation from quantum mechanics, I have tried putting log on both sides and pull the powers down, but I'm not sure whether I'm doing the correct thing, please help me. Thanks!!
For simplicity, let's just call a0 = a for now.
So:
P(r) = r^2*R(r)^2 = r^2*[(2 - r/a)^2 * e^{-r/a}] / (2a)^3
Now use the product and chain rules:
P'(r) = 2r*[(2 - r/a)^2 * e^{-r/a}] / (2a)^3 + r^2*[2(2 - r/a) * (-1/a) * e^{-r/a}] / (2a)^3 + r^2*[(2 - r/a)^2 * e^{-r/a} * (-1/a)] /(2a)^3
Now we need to simplify a bit.
P'(r) = 2r*[(4 - 4r/a + r^2/a^2) * e^{-r/a}] / (2a)^3 + r^2*[(-4/a + 2r/a^2) * e^{-r/a}] / (2a)^3 + r^2*[(-4/a + 4r/a^2 - r^2/a^3) * e^{-r/a}] /(2a)^3
Now factor a e^{-r/a}/(2a)^3 and do some further multiplying:
P'(r) = {8r - 8r^2/a + 2r^3/a^2 + -4r^2/a + 2r^3/a^2 - 4r^2/a + 4r^3/a^2 - r^4/a^3} * e^{-r/a}/(2a)^3
Now collect powers of r:
P'(r) = {-(1/a^3)r^4 + (8/a^2)r^3 - (16/a)r^2 + 8r} * e^{-r/a}/(2a)^3
Now make things nice with that (2a)^3 at the end:
P'(r) = {-(1/(8a^6))r^4 + (1/a^5)r^3 - (2/a^4)r^2 + (1/a^3)r} * e^{-r/a}
-Dan
Oh, ok, i guess i had to have some knowledge of quantum mechanics to answer this question. i thought R(t) was a typo or something, i never knew R(t) was different from P(t) and you had to relate them by another function. Luckily there's a physics maestro on this site!
I almost didn't see it at first either, but P(r) = [rR(r)]^2P(r) = [rR(r)]^2,
R(r) =
Find P'(r).
This is an equation from quantum mechanics, I have tried putting log on both sides and pull the powers down, but I'm not sure whether I'm doing the correct thing, please help me. Thanks!!
If you had noticed that, you would likely have gotten the correct derivative.
O wow, you know i still didn't see that until just now! I thought topsquark pulled that formula out of his head.
i seriously think i need to relearn how to read
but still, the font size for that thing was like 8 while for the other function its like 500, you won't notice that if your itching to differentiate something like i usually am
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