Question 5

[M.C.]

Dr. Ferdinand,
I was thinking that the intensity of light does not change the rate of elecrton emission, but it is photon energy which does that. However, here, the intensity is changing the rate. Could you please explain this alittle bit more.
M.C.

[admin]

Intensity, amplitude and energy are all positively correlated. Thus I see no difference between the intensity of light and photon energy.

[Matthew]

On the MCAT, if question asks about sound and relation with intensity use the formula (I=P/A) You can think about this conceptually by speaking loud next to someone compared to someone far away. The "someone" far away doesn't get the same intensity, it's much lower than they poor fella next to you why you are yelling.

On the other side of the grass, if on the MCAT talks about light and intensity, remember that the square of Amplitude is equal to the intensity of the light. And as Dr. Ferdinand so clearly pointed out, when you think intensity for light, it is the same as energy.

[kazuma]

quick question, is the intensity related to energy (amplitude) bc of hookes law? 1/2kA^2? bc intensity is power/area, and power is just energy/time. I know all waves are analogous to an oscillating spring.

[admin]

quick question, is the intensity related to energy (amplitude) bc of hookes law? 1/2kA^2?

Seems reasonable.

[kazuma]

. A sound wave from a stationary observer is bounced off an object approaching with velocity v. The reflected wave, relative to original wave, has:

1. increased frequency and same velocity. [x]
2. decreased frequency and increased velocity. [x]
3. same frequency and increased velocity. [x]
4. same frequency and same velocity. [x]

CORRECT:
Your Answer: A
Correct Answer: A
Explanation
The speed of the reflected sound wave is the same since it depends on the medium of transmission. Its frequency however has changed according to the Doppler effect. The object which reflects the sound wave can be viewed as a sound source approaching with velocity ; recall from memory or experience that when the source and detector move toward each other with a positive net velocity, the frequency of the emitted wave (mechanical or electromagnetic) is increased.


one thing about the explanation; the doppler effect of an electromagnetic wave is dependent on the relative speeds of the source and detector right (unlike that of a transverse wave)? so if they are both moving at the same speed and in the same direction, there is no doppler effect??

[admin]

That's correct.

PS: Relativity is not covered in the MCAT.

[bjpatel2178]

I got this one right also, but had a question.


f is doppler shifted, but if one wave is reflected and is going the opposite direction then its speed should also be increased (based on the frame of reference we are dealing with).

If you think about it closely, or at least the way I interpreted it the speed of the reflected wave RELATIVE to the speed of the original wave, both in opposite directions should be higher. If it was relative to the stationary frame of reference it would be the same.

Does this make sense? Can you help me on this Dr. Ferdinand?

Thanks!

[quadalpha]

Did the order of the question change? Anyway, this is about Be ionization:

The reason most people would rule out D is because whether ionization happens is independent of intensity. It's much more reasonable to imagine that the increase in intensity is non-linear than to suppose there must be a wait for a "tiny fraction of a second", which is by no means guaranteed to be linear anyway.[/list]

[hoanle888753]

I still have some difficulty in understanding how increasing the light intensity can affect the rate of electron emission in question 5. I thought that an electron can only be ionize at a specific energy level and thus specific frequency of light. Ques.1 seem to agree with the idea that intensity should have no affect on the emission of electron since it state that "no matter how much red light was shown on the metal light, no electrons was detected." It appear to me that increasing the intensity of light doesn't mean that the energy of light has increase (since the light still have the same specific frequency), but rather the rate at which the energy is deliver has increase. Thus the rate at which the electron is emitted increase with the intensity only if the light energy to begin with correspond to that of the electron energy, but if the light have less energy that that of the electron, then no matter how great the intensity is, the electron will not be emitted. Does this make sense? Thanks

[jellywing_2058]

Yes, it does make sense. The electrons require some minimum amount of energy to be ejected out of the beryllium atoms; this minimum energy is called the Work Function or the Threshold Energy, and requires radiation of a minimum frequency to provide this much energy.

W = h f W = Work Function, f = Minimum frequency
It is obvious that below this frequency there will be no emission.

However, increasing the frequency f will mean more energy carried by the light photons, and therefore more energy imparted to the electrons and therefore electrons emitted will have more kinetic energy.
Increasing the Intensity implies that the number of Photons per second have increased; more photons are now impinging per second and therefore more electrons are now being emitted. They are thus linearly or directly proportional.

20 photons/ second result in an R = 20 electrons/ second. We are of course assuming that each incident photon will eject an electron.

Doubling the intensity would mean that the number of Photons per second has now become 40 photons/ second causing an emission of 40 electrons per second, that is a R = 40 electrons/ second. Assuming again, of course that each incident photon ejects an electron.

The Basic condition of course remains that the Frequency of the incident photons is above the Threshold Frequency f, where h f = W, W being the minimum energy required to cause emission and h, the Planck’s Constant = 6.626 x 10^-34 J s.

I hope this helps