Hhn
You would use Planck's constant which is 6.626*10^-34 J*s and multiply by the frequency given and the answer is the as you said.
Yes - that's how it works.
mc2
i have the same question on my test haha for me, the answers are: a) The number of electrons ejected per second b) the maximum kinetic energy of the ejected electrons c) the threshold frequency of the ejected electrons d) the time lag between the absorption of blue light and the start of emission of the electrons e) none of these A the number of electrons ejected per second,,,,, correct answer
In any circumstance where a threshold of energy is required to free an electron from a bound state, an incoming photon must have at least that energy to do the job. The energy of a photon is proportional to the frequency of the light, so the minimum energy corresponds to a minimum frequency of the light, or maximum wavelength necessary to free an electron. This observation was a major step in the development of radiation theory (Einstein).
The threshold frequency is the minimum frequency of radiation required to raise the potential energy of the most energetic electrons in a metal to zero, therefore giving the free electron (once emitted) a velocity of zero. The work function of a metal surface is the energy required to remove the most energetic electron from it. Each metal has a different work function, with the negative of this the maximum potential of the de-localised electrons in the metal. W=hf0 The formula: Kmax = hf - W (in Joules) gives the maximum kinetic energy of the electron. As h is Planck's constant, and f is the frequency provided to the metal (ie. frequency of the electromagnetic wave that had irradiated the metal), and W is the Work Function of the metal, if the frequency provided to the metal is the threshold frequency, then the electron has kinetic energy of zero. However, if more than the threshold frequency is provided, the electron will have a kinetic energy > 0. I hope that's explained okay! Bec
The greatest wavelength of radiation for a specified surface for the emission of electrons.
Threshold frequency (fo) is minimum frequency at which electrons are ejected from a metal.
w=hf w-work funtion h-constant f-threshold frequency the work funtion is the minimum energy required to remove the electrons on the metal
9.39*10^14 Hz is the threshold frequency of cesium.
It doesn't, and that's the whole big mysterious fact about the photoelectric effect that was standing Physics on its ear about 100 years ago. It doesn't matter how bright the light is, there's no photoelectric effect if the light is below the threshold frequency. And if it's above the threshold frequency, it doesn't matter how dim the light is, those electrons come streaming off of the surface of the target.
The threshold frequency for photoelectric emission is the smallest possible frequency a photon can have to be absorbed/emitted by an electron moving between energy levels in an atom. Explanation: Since electrons can't exist /between/ energy levels, and each electron would be moved a very specific amount by any given photon, only photons of certain frequencies can be properly absorbed/emitted, necessitating a minimum frequency.
You would use Planck's constant which is 6.626*10^-34 J*s and multiply by the frequency given and the answer is the as you said.
Yes - that's how it works.
non-threshold
The electromagnetic force between the electrons in the person's atoms and the electrons in the wall's atoms is greater than the force of a person pushing against it. The atoms in the wall are connected via their electrons with a certain amount of force; if the force applied to the wall exceeds this threshold, then the object acting on the wall will go through it.
The photoelectric current is directly proportional to intensity.It also depends upon frequency, but frequency more than "THRESHOLD FREQUENCY" does not effect the current.The no. of electrons emitted per second by a photo-sensitive surface is directly proportional to the intensity of the incident radiations.So,the photoelectric current depends upon the intensity of the incident radiations.