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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.
Calculate the de Broglie wavelength of the electrons emitted from a copper crystal when irradiated by photons from the first line and the sixth line of the Lyman series.
`h=6.6256xx10^(-34) J s," "m_(e)=9.1091xx10^(31) kg," "c=2.99792xx10^(8) m s^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
The wavelength of an electron is:
`lambda=h/p=(h)/sqrt(2E_("kin")m_(e))`
`(p=m_(e)v_(e) "and" E_("kin")=p^(2)/(2m_(e)))`
`DeltaE_(2 rarr 1): lambda_(1)=4.16xx10^(-10) m=4.16 Å`
`DeltaE_(7 rarr 1): lambda_(2)=5.20xx10^(-10)m =5.20 Å`
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