I Wanna Be The Forums!

Please login or register.

Login with username, password and session length
Advanced search  

News:

Gaiden 1.2 patch released. Click here to download!

Pages: 1 2 3 [4]

Author Topic: math(s) thread  (Read 7231 times)

SilentLoner

  • Master of Time and Space
  • Hero Member
  • *****
  • Posts: 734
    • View Profile
Re: math(s) thread
« Reply #45 on: January 30, 2012, 06:33:13 pm »

That's a horrible idea, and I'm disappointed in you for coming up with such an idea in the first place.
Logged

Dagnarok

  • Neutral Good Moderator
  • Administrator
  • The Guy
  • *****
  • Posts: 3685
  • Commander of the USP Talon
    • View Profile
Re: math(s) thread
« Reply #46 on: January 30, 2012, 06:36:48 pm »

It's your funeral. :P
Logged
Quote from: Tilyami
Shuu is like the monkey's paw if the monkey's paw was a partridge.

Benxamix2

  • Sr. Member
  • ****
  • Posts: 387
  • INACTIVE
    • View Profile
Re: math(s) thread
« Reply #47 on: January 30, 2012, 06:37:42 pm »

* Benxamix2 cries
Logged

Neko Desu

  • The Guy
  • *****
  • Posts: 1898
    • View Profile
Re: math(s) thread
« Reply #48 on: January 30, 2012, 11:29:15 pm »

By the way, I've been stuck on this physics question for a while now.

When light of wavelength 2.08 * 10^-7 m falls on a photosurface, a voltage of 1.40 V is required to stop the emitted electrons from reaching the anode. What is the largest wavelength of light that will result in the emission of electrons from this photosurface?

I was thinking since V = E/Q then E = 1.40 * 1.6 * 10-19 which gives E = 2.24 * 10^-19 J, then sub in this value for E = hfc and since h is just planck's constant then you just solve for the critical frequency. Then you use λ = c / f where 'c' is the speed of light and 'f' is the critical frequency, resulting in a wavelength of 8.9 * 10^-7 m, but the answer in my textbook says it's 2.7 * 10^-7 m

halp
Logged

Kwiftee

  • R ANGRY
  • Sr. Member
  • ****
  • Posts: 491
  • That guy who's always ranting about stuff.
    • View Profile
    • http://no
Re: math(s) thread
« Reply #49 on: January 31, 2012, 01:49:54 am »

By the way, I've been stuck on this physics question for a while now.

When light of wavelength 2.08 * 10^-7 m falls on a photosurface, a voltage of 1.40 V is required to stop the emitted electrons from reaching the anode. What is the largest wavelength of light that will result in the emission of electrons from this photosurface?

I was thinking since V = E/Q then E = 1.40 * 1.6 * 10-19 which gives E = 2.24 * 10^-19 J, then sub in this value for E = hfc and since h is just planck's constant then you just solve for the critical frequency. Then you use λ = c / f where 'c' is the speed of light and 'f' is the critical frequency, resulting in a wavelength of 8.9 * 10^-7 m, but the answer in my textbook says it's 2.7 * 10^-7 m

halp

Well, we know that the stopping voltage is 1.40V. This corresponds to an energy of 2.24*10^-19 J, as you said. That energy represents the maximum possible kinetic energy that an emitted photoelectron can have, when light of the given wavelength is used.

Now, we use EK max = hf - W, in order to get the work function. Or, in this case, we use E = hc/λ -W
2.24*10^-19 = (6.63*10^-34) * (3.0*10^8) / (2.08*10^-7)   -  W.
solving for W gives W = 7.3225*10^-19 J.

Now, we want the cutoff wavelength; the point at which EK max = 0. (ie. electrons aren't really emitted at all.) Hence, hc/λ - W = 0
W = hc/λ
λ = hc/W

so, λ = (6.63*10^-34) * (3.0*10^8) / (7.3225*10^-19)
 which is 2.7*10^-7 m.


The mistake you were making is that the energy you had found was NOT the work function. Rather, it was the amount of energy that was left over once you had subtracted the work function from the energy of the given photons.
If that doesn't make sense I'll try and explain it again; these types of problems were always the most difficult we got in our physics exams. You had to make sure you were visualising what type of energy you were referring to correctly. Talking of which; if you know how to use electron volts as your units rather than joules, it makes it a fair bit simpler.
« Last Edit: January 31, 2012, 01:51:28 am by Kwiftee »
Logged
zyxfrlpso
Quote from: Yuletideriz
My god, if I could shit gold and I give it to you, you would still bitch because it was technically crap.
how about you GO ON IRC RIGHT NOW PLEASE

Neko Desu

  • The Guy
  • *****
  • Posts: 1898
    • View Profile
Re: math(s) thread
« Reply #50 on: January 31, 2012, 01:54:08 am »

Thanks a lot!
Logged

Kwiftee

  • R ANGRY
  • Sr. Member
  • ****
  • Posts: 491
  • That guy who's always ranting about stuff.
    • View Profile
    • http://no
Re: math(s) thread
« Reply #51 on: February 04, 2012, 07:50:42 pm »

for x > 0
x = 1 + 1 + 1 + ... + 1; } x times
x2 = x + x + x + ... + x; } x times

d/dx(LHS) = d/dx(x2) = 2x.

d/dx(RHS) = d/dx(x + x + x + ... + x); } x times
              = d/dx(x) + d/dx(x) + d/dx(x) + ... d/dx(x); } x times
              = 1 + 1 + 1 + ... + 1; } x times
              = x.

but 2x ≠ x.
what
Logged
zyxfrlpso
Quote from: Yuletideriz
My god, if I could shit gold and I give it to you, you would still bitch because it was technically crap.
how about you GO ON IRC RIGHT NOW PLEASE

Neko Desu

  • The Guy
  • *****
  • Posts: 1898
    • View Profile
Re: math(s) thread
« Reply #52 on: February 14, 2012, 10:28:03 pm »

HOMEWORK, HALP

The index of refraction for blue light of wavelength 4.5 * 10-7 for a particular kind of glass is 1.328 and for red light of wavelength 6.5 * 10-7 it is 1.321. White light is incident on an equilateral triangular prism made of this glass parallel to one of its bases. What are the angles that the blue and red rays make with the normal as they emerge from the prism?
Logged
Pages: 1 2 3 [4]