Calculating Electron Emission from Lithium with 100W Visible Light

[silverek]

The wavelength of visible light is between 400nm and 760nm. We shed a 100W visible light on a piece of metal.

In classical electrodynamics, what is the number of electrons escaping from the Lithium surface per second, if the metal absorbs 20% of the light? Lithium has a work function of 2.38eV.

 

[Gokul43201]

What minimum wavelength will lead to photoelectron emission (determine this from the work function) ? Now assume that the visible light has a flat frequency distribution and determine the number of sufficiently energetic photons incident on the metal (from the relevant fraction of total power). 20% of this number of photons may be assumed to each produce one photoelectron.

 

[R0man]

The number of electrons escaping from the surface of Lithium per second can be calculated using the formula:

Number of electrons per second = (Power of light * Absorption coefficient * Surface area of metal) / (Planck's constant * Speed of light * Work function)

Substituting the given values, we get:

Number of electrons per second = (100W * 0.2 * Surface area of metal) / (6.626 x 10^-34 J*s * 3 x 10^8 m/s * 2.38 eV)

To determine the surface area of the metal, we need to know the dimensions of the piece of metal. Assuming it is a square with sides of 1 cm, the surface area would be 1 cm^2 or 1 x 10^-4 m^2.

Plugging in the values, we get:

Number of electrons per second = (100W * 0.2 * 1 x 10^-4 m^2) / (6.626 x 10^-34 J*s * 3 x 10^8 m/s * 2.38 eV)

Simplifying, we get:

Number of electrons per second = 5.28 x 10^17 electrons/s

Therefore, if the metal absorbs 20% of the 100W visible light, the number of electrons escaping from the surface of Lithium per second would be approximately 5.28 x 10^17 electrons.