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UV light of wavelength 1800Å is incident...

UV light of wavelength `1800Å` is incident on a lithium surface whose threshold wavelength `4965Å`. Determine the maximum energy of the electron emitted.

Text Solution

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The wavelength of incident light `= 1800 Å`
The threshold wavelength `lambda_(0) =4965Å`
`K_("max") = hr-hr_(0)`
`= ( hc)/( lambda) - ( hc)/( lambda_(0))`
`K_("max") = hc[ (1)/( lambda) - (1)/( lambda_(0))]`
=hc
`= 6.626 xx 10^(-34 ) xx3 xx 10^(8) [ (1)/( 1800) - (1)/( 4965)]`
` = 19.86 xx 10^(-26) [ ( 4965-1800)/( 8937xx 10^(3))] xx ( 10^(-10))/( 10^(-20))`
`= 19.86 xx 0.3541 xx 10^(-19) = 7.03 xx 10^(-19) J`
`= ( 7.03 xx 10^(-19))/( 1.6 xx 10^(-19)) =4.40 eV`
`K_("max") = 4.40 eV`
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