Question

What is the wavelength of light that hits potassium metal in order to eject photoelectrons with a wavelength of `"0.57 nm"`? The ionization energy of potassium is `"4.34 eV"`.

Answer 1

I got ultraviolet light of about `"138.24 nm"`. If you got `"0.57 nm"`, you assumed the photoelectrons had no energy and that the threshold energy was based on the `"0.57 nm"` wavelength, which is a possible but nonphysical mistake.

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We first consider the photoelectric effect.

As we are bombarding a metal surface to eject an electron, it has leftover kinetic energy `K_e` equal to the difference between the incoming energy `hnu` and the threshold energy `phi = hnu_0` that was overcome:

`K_e = hnu - hnu_0 = hnu - hnu_0`

where `h` is Planck's constant.

Recall that for every wavelength `lambda` that light, with speed `c = 2.998xx10^(8) "m/s"`, cycles through, we speak of a frequency `nu`. Thus, `nu = c/lambda`, and:

`K_e = (hc)/(lambda) - hnu_0`

Since the photoelectrons are to have the wavelength of `"0.57 nm"`, having mass, they follow the de Broglie relation:

`lambda = h/(mv) = h/p`

where `m`, `v`, and `p` are mass, speed, and momentum, respectively.

Recall that for mass-ive particles, `K_e = 1/2mv^2 = p^2/(2m)`. Thus,

`K_e = (h//lambda)^2/(2m)`

`= (6.626 xx 10^(-34) "kg"cdot"m"^2cdot"s"^(-1) // 0.57 xx 10^(-9) "m")^2/(2 xx 9.109 xx 10^(-31) "kg")`

`= 7.417 xx 10^(-19) "J"`,

or `"4.63 eV"`, is the energy of the photoelectrons leaving potassium metal. The actual ionization energy of potassium is `"4.34 eV"`, which means that based on that threshold energy, we have:

`overbrace("4.63 eV")^(K_e) = (hc)/lambda - overbrace("4.34 eV")^(hnu_0)`

or

`overbrace(7.417 xx 10^(-19) "J")^(K_e) = (hc)/lambda - overbrace(6.952 xx 10^(-19) "J")^(hnu_0)`

Thus,

`color(blue)(lambda) = (hc)/((7.417 + 6.952) xx 10^(-19) "J")`

`= (6.626 xx 10^(-34) "J"cdot"s" cdot 2.998 xx 10^(8) "m/s")/(1.437 xx 10^(-18) "J")`

`= 1.382 xx 10^(-7) "m"`

`=` `color(blue)("138.24 nm")`