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Light of wavelength `1824 Å`, incident on the surface of a metal , produces photo - electrons with maximum energy `5.3 eV`. When light of wavelength `1216 Å` is used , maximum energy of photoelectrons is `8.7 eV`. The work function of the metal surface is
A. `3.5 eV`
B. `13.6 eV`
C. `6.8 eV`
D. `1.5 eV`

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Correct Answer - D
`E = W_(0) + K_(max)`. "From the given data" `E is 6.78 eV`
( for `lambda = 1824 Å)` or `10.17 ev ( for lambda = 1216 Å)`
`:. W_(0) = E - K_(max) = 6.78 - 5.3 = 1.48 eV`
or `W_(0) = 10.17 - 8.7 = 1.47 eV`.

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