Question

First ionization energy of hydrogen atom= 2.18 aJ (attojoules). What the frequency and wavelength, in nanometers, of photons capable of just ionizing hydrogen atoms?

Assuming an ionization efficiency of 76.0%, how many such photons are needed to ionize `1.00 xx 10^16` atoms?

Answer 1

`nu = 3.29 xx 10^15 "s"^(-1)`

`lambda = "91.1 nm"`

`1.32 xx 10^16 "photons"`

`E_"LASER" = "0.0287 J"`

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There are `10^18` atto whatevers in one of those. So, `"2.18 aJ" = 2.18 xx 10^(-18) "J"`.

The energy of a hydrogen atom is given by:

`E_n = -2.18 xx 10^(-18) "J" cdot 1/n^2`

where `n` is the quantum number for the energy level. `n = 1, 2, 3, . . .`.

From this we get the Rydberg equation for when the electron moves from one energy level to another:

`DeltaE = -2.18 xx 10^(-18) "J" (1/n_f^2 - 1/n_i^2)`

where `f` indicates the destination and `i` indicates the starting point.

If you want to remove an electron completely, it is the limiting case where `n_f -> oo`, because the electron moves up in energy until it escapes the atom.

So the ionization energy is:

`DeltaE_(1->oo) = DeltaH_"IE"`

`= -2.18 xx 10^(-18) "J" cdot (1/oo^2 - 1/1^2)`

`= 2.18 xx 10^(-18) "J"`

This ionization was caused by one photon (i.e. a packet of light), so this photon would need frequency `nu` and wavelength `lambda` to cause that, according to

`E_"photon" = DeltaE_(1->oo) = hnu = (hc)/lambda`

where `h = 6.626 xx 10^(-34) "J"cdot"s"` is Planck's constant and `c = 2.998 xx 10^8 "m/s"` is the speed of light.

As a result,

`color(blue)(nu) = (DeltaE_(1->oo))/h = (2.18 xx 10^(-18) "J")/(6.626 xx 10^(-34) "J"cdot"s")`

`= color(blue)ul(3.29 xx 10^15 "s"^(-1))`

`color(blue)(lambda) = (hc)/(DeltaE_(1->oo)) = (6.626 xx 10^(-34) "J"cdot"s" cdot 2.998 xx 10^8 "m/s")/(2.18 xx 10^(-18) "J")`

`= 9.11 xx 10^(-8) "m"`

`=` `color(blue)ul("91.1 nm")`

If you then want to ionize `bb(76%)` efficiently and manage to nail `1.00 xx 10^16` atoms, it would mean that `1.00 xx 10^(16)` is `76%` of the atoms you make this attempt on, meaning `76%` of the number of photons you supply.

Therefore, the number of photons you ACTUALLY need to supply is:

`(1.00 xx 10^(16))/(0.76) = color(blue)(1.32 xx 10^16 "photons")`

Recall that for every photon that we use in this scenario, it has energy:

`E_"photon" = 2.18 xx 10^(-18) "J/photon"` against hydrogen atom

Therefore, for a bunch of photons, we need maybe a few hundredths of joules.

`color(blue)(E_("LASER")) = (2.18 xx 10^(-18) "J")/cancel("1 photon") xx 1.32 xx 10^16 cancel"photons"`

`=` `color(blue)ul("0.0287 J")`

And by the way, we have just stated that this laser we are using contains `1.32 xx 10^16` photons, and that `76%` of those photons hit the sample of hydrogen atoms perfectly to ionize them all.