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?

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

#lambda = "91.1 nm"#

#1.32 xx 10^16 "photons"#

#E_"LASER" = "0.0287 J"#

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.