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When a surface is irradiated with light ...

When a surface is irradiated with light of wavelength `4950 Å`, a photocurrent appears which vanishes if a retarding potential greater than 0.6 volt is applied across the phototube. When a second source of light is used, it is found that the critical potential is changed to 1.1 volt.
The work- function of the emitting surface is i

Text Solution

Verified by Experts

Data: `lambda=4950 Å = 4.95 xx 10^(-7) m, V_(s) = 0.6V`
`KE_("max") =V_(S)e = hv-phi = (hc)/lambda - V_(s)e` (in joule)
`=(hc)/(elambda)- V_(s)` (in electronvolt)
`=((6.63 xx 10^(-34))(3 xx 10^(8)))/((1.6 xx 10^(-19))(4.95 xx 10^(-7)))-0.6`
`=19.89/(7.92) - 0.6 = 2.512 - 0.6 = 1.912` eV
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