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The energy of stable states of the hydro...

The energy of stable states of the hydrogen atom is given by `E_(n)=-2.18xx10^(-8)//n^(2)[J]` where n denotes the principal quantum number.
Can a single photon, emitted in the first and//or sixth line of the Lyman series, ionize:
{:a) another hydrogen atom in its ground state?
{:b) a copper atom in the Cu crystal?
The electron work function of Cu is `phi_(Cu)=7.44xx10^(-19) J`.

Text Solution

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The correct Answer is:
{:a) The ionisation energy is equal to `DeltaE_(n rarr 1)=2.18xx10^(-18) J`. Both `DeltaE_(2 rarr 1)` and `DeltaE_(7 rarr 1)` are smaller than `DeltaE_(i rarr 1) ` and a single photon emitted from these transitions is not able to ionize a hydrogen atom.
{:b) Ionisation of copper in a Cu-crystal is related to the photoelectric effect:
`hv=phi_(Cu)+E_("kin")=phi_(Cu)+`^(1)//_(2) m_(e)V^(2)``
Because `DeltaE_(2 rarr 1) gt phi_(Cu)` and `DeltaE_(7 rarr 1) gt phi_(Cu)` both photons are indeed able to ionize a cu-atom in the crystal. The kinetic energy of the emitted electrons is:
`DeltaE_("kin")(2 rarr 1)=DeltaE_(2 rarr 1)-phi_(Cu)=8.91xx10^(-19)J`
`DeltaE_("kin")(7 rarr 1)=DeltaE_(7 rarr 1)-phi_(Cu)=13.91xx10^(-19) J`
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