when a clean surface of potassium metal is exposed to blue light, electrons are emitted.?

Question

if the intensity of the blue light is increased, the _________ will also increase.

a. maximum kinetic energy of the ejected electrons
b. number of electrons ejected per second
c. time lag between the absorption of blue light and the start of emission of the electrons
d. threshold frequency of the ejected electrons

Answer 1

This is called the "Photoelectric Effect", and is true of many materials, particularly metals. (this term was coined by Nikkola Tesla in 1901...)

Classical mechanics predicts that both the number, and the kinetic energy of the ejected electrons should increase. However, in practice, only the *number* of ejected electrons increases.

Einstein explained the correct behavior of "Photo-electrons" in 1905, for which he won the Nobel Prize in physics in 1921.

Light has properties of both a wave, and a particle. Light can be "quantized" as individual "wave packets", or "photons." Each photon has a set amount of kinetic energy, depending only on it's frequency:

E = h*f;

where h is "Plank's constant."

Increasing the intensity of the light only increases the *number* of photons, not their individual energy. Thus, the maximum kinetic energy of ejected electrons is roughly proportional only to the *frequency* of the incoming light, not it's intensity.

In general, one photon causes only one electron to be ejected. The electron has to have at least a certain amount of kinetic energy, in order to escape binding forces in the base material. Thus, below a certain "threshold frequency", light simply does not have enough energy to cause electrons to escape. The kinetic energy of the ejected electrons is given by:

½ me*v² = h*f - h*fo ;

where me is the mass of an electron, v is their velocity, h is Plank's constant, f is the frequency of the incoming light, and fo is the cutoff frequency of the material in question.

Hope that is not too confusing,
~Soylent Yellow

Answer 2

easily, it is any incorrect way around: The ability of the photons extraordinary the exterior is given by potential of: Ephotons = hf the place h = 6.626x10^-34 Js (Planck's consistent) and f is the frequency. for this reason, larger-frequency photons carry extra ability. whilst larger-ability photons hit the exterior, the exterior emits larger-ability electrons in accordance to the equation: Eelectrons = hf - ? the place ? is the "paintings function" of the metallic (i.e. the minumum ability required to eject an electron). once you strengthen the mild intensity, you strengthen the style of photons incident on the exterior, and for this reason, the style of electrons ejected from the exterior. that's the "photoelectric result", and it paved the way for quantum mechanics (besides as a Nobel prize for Einstein).

Answer 3

This is the Photoelectric effect I believe.....

Answer is B, because.....

K.E. of Electron = hf - work function

Changing intensity doesn't give photoelectrons more energy, it just increases the number ejected.