homer's physics corner

[Homer]

another question similar to that one.

Light of wavelength 600 nm falls onto the metallic surface. the retarding potential for the photoelectron is 0.24V. Find the work function of the metal. answer: 1.83eV

[SocialRhubarb]

Cut off potential is 6.8V. This means that this is the point at which the voltage is just high enough to stop all the photoelectrons ejected.

Volts are Joules per coulomb. One electron is .

The voltage is sufficient so that the photoelectrons with the maximum amount of kinetic energy are just stopped. This means that the energy lost by the electron as it passes through this voltage is equal to its maximum kinetic energy, since in this case, this is the potential at which the electrons are just stopped. I emphasise this point because this isn't the case for all questions like this, and some, like nliu says, involve work functions etc.

But, if 6.8V is just enough to stop the most energetic photoelectrons, the maximum energy a photoelectron can have is:

[lzxnl]

Oh, so he means in the event that we remove the voltage and that we pre-suppose the light-shining is already taking place. OK.

[Homer]

another question similar to that one.

Light of wavelength 600 nm falls onto the metallic surface. the retarding potential for the photoelectron is 0.24V. Find the work function of the metal. answer: 1.83eV

^this one
and
according to some notes I have when you increase the intensity of the light source you get more current, but when you increase the frequency you get the same current. i though more frequency=more enrgy=more current? I am still having trouble understanding the effect of frequency, intensity and wavelength upon photoelectric current.

[Conic]

Each photon displaces one electron. Increasing the intensity increases the number of photons, and hence increases the number of electrons (ie, higher current). Increasing the frequency increases the energy of the photon, which means more energy is transferred to the electron (ie, it has more kinetic energy). Increasing the wavelength has the same effect as decreasing the frequency (as ).

Edit: answer other question

  and 



(W is the work function, h is Planck's constant, f is the frequency, eV0 is the stopping voltage)

The frequency of the light can be found using :

Now we can find the work function:

[Homer]

In a double slit interference experiment, an electron beam travels through two narrow slits, 20mm apart, in a piece of copper foil. The resulting pattern is detected photographically at a distance of 2m. The speed of electron is 0.1% of the speed of light.
Q) what do you expect to see on the photographic plate?

when i calculates the de Broglie wavelenght and divided it by 20x10^-3 I got a number way less than 1, so wouldn't that mean that there would be no differaction or interference?

Answer says: A series of bright and dark fringes :/

[Homer]

Regarding emission and absorption spectrum
when a beam of electron is directed towards an atom,
why is it that, no matter what energy an electron carries (given that it is less than the energy required to ionise the atom), the atom can be excited, whereas when photons are directed only specific energy is required to excite the atom? :s

[~T]

Regarding emission and absorption spectrum
when a beam of electron is directed towards an atom,
why is it that, no matter what energy an electron carries (given that it is less than the energy required to ionise the atom), the atom can be excited, whereas when photons are directed only specific energy is required to excite the atom? :s

Electrons transfer their energy by 'collisions', just like how you can bump into someone and move them by transferring some of your kinetic energy. Photons (in this case) cease to exist in a transfer of energy with electrons. In essence, they must give all their energy or none of their energy. It is more useful to think of photons as packets of energy (in this case), not as particles. Essentially, as long as the incoming electron has enough energy to move an electron to the first excited energy level, then a collision may do so.

In a double slit interference experiment, an electron beam travels through two narrow slits, 20mm apart, in a piece of copper foil. The resulting pattern is detected photographically at a distance of 2m. The speed of electron is 0.1% of the speed of light.
Q) what do you expect to see on the photographic plate?

when i calculates the de Broglie wavelenght and divided it by 20x10^-3 I got a number way less than 1, so wouldn't that mean that there would be no differaction or interference?

Answer says: A series of bright and dark fringes :/

You've divided wavelength by the slit *spacing*. The ratio to find the qualitative degree of diffraction is the wavelength divided by the slit *size*. We are just told that they are narrow, so we presume they are narrow enough to cause diffraction and result in an interference pattern.

[Homer]

1)What happens when white light undergoes young's double slit experiement (diffraction/interference)? Is it the same as a normal coherent light source?

2) suppose when calculating the standing waves of a hydrogen atom, you get a answer as n=3. What does that mean? Does it mean the third excitation state or the second? and why?

[lzxnl]

1)What happens when white light undergoes young's double slit experiement (diffraction/interference)? Is it the same as a normal coherent light source?

2) suppose when calculating the standing waves of a hydrogen atom, you get a answer as n=3. What does that mean? Does it mean the third excitation state or the second? and why?

1. White light consists of a multitude of frequencies. You'll get a bright maximum in the middle where the path difference is zero, but then different frequencies interfere destructively and constructively at different path differences. I don't know what you'll get, but you'll get a coloured mess.

2. That is the second excitation state. The lowest energy state is n=1; the energy formula is -13.6/n^2 in eV.

[Applebottom]

I think the book is referring only to when they are moving down, and not up. But yeah, it could have been worded better.

If she was going up acceleration would still be 9.8 ms^-2 down but her velocity would be up. Therefore the normal force should be zero because the only force acting on the toy and the girl is gravity after her feet has left the trampoline.

[Applebottom]

1. White light consists of a multitude of frequencies. You'll get a bright maximum in the middle where the path difference is zero, but then different frequencies interfere destructively and constructively at different path differences. I don't know what you'll get, but you'll get a coloured mess.

2. That is the second excitation state. The lowest energy state is n=1; the energy formula is -13.6/n^2 in eV.

In the text book page 417 there is a diagram of white light passing through a diffraction grating which is in theory just loads of double slits.
The picture shows a bright band in the middle that shows the rainbow spreading from read to blue from the center band, away in both directions. the rainbow repeats itself and gets lighter the further away it is from the dark band.
In a double slit the diffraction is not as clear but it can be assumed that it would attempt to follow this model.
We performed the experiment in class and I remember seeing the rainbow come out from red to blue in this way from every bright band when looking at white light.