Answer:
Kinetic energy of photoelectron = \[e{{V}_{0}}\]
where, \[{{V}_{0}}\] = stopping
potential; e = charge of electron
\[\therefore \] \[K.E.=1.6\times
{{10}^{-19}}\times 0.35=5.6\times {{10}^{-20}}\text{J}\]
Absorbed energy by electron = \[\frac{hc}{\lambda
}\]
\[=\frac{6.626\times
{{10}^{-34}}\times 3\times {{10}^{8}}}{256.7\times {{10}^{-9}}}\]
\[=7.743\times
{{10}^{-19}}\text{J}\]
Absorbed Threshold energy Kinetic energy
of
energy =
or + photoelectron
Work
function
\[7.743\times {{10}^{-19}}={{W}_{0}}+5.6\times
{{10}^{-20}}\]
\[{{W}_{0}}=7.183\times
{{10}^{-19}}\text{J}\]
\[=\frac{7.183\times
{{10}^{-19}}}{1.6\times {{10}^{-19}}}eV=4.48eV\]
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