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A metallic surface when illuminated with light of wavelength 3333 Å emits electrons with energies upto 0.6 eV. Calculate the work function of the metal. Data : lambda=3333 Å, K.E=0.6eV, W=?

Answer»

SOLUTION :Work function, W = hv - kinetic energy
or `W=(HC)/(lambda)-K.E`
`=((6.26 xx 10^(-34) xx 3 xx 10^(8))/(3333xx10^(-19)))-(0.6xx1.6xx10^(-19))`
`=(5.96 xx 10^(-19))-(0.96 xx 10^(-19))`
`W=5 xx 10^(-19)J`
`W=(5 xx 10^(-19))/(1.6 xx 10^(-19))eV.`
W = 3.125 eV.


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