A) 4.20 eV
B) 0.15 eV
C) 2.10 eV
D) 1.05 eV
Correct Answer: D
Solution :
From Einsteins equation for photoelectric effect, we have \[{{E}_{k}}=W+e{{V}_{0}}\] where \[{{E}_{k}}\] is maximum kinetic energy, W is work function. Also \[{{E}_{k}}=\frac{hc}{\lambda }\] \[\therefore \] \[\frac{hc}{350\times {{10}^{-9}}}=W+2e{{V}_{0}}\] ...(1) \[\frac{hc}{540\times {{10}^{-9}}}\,=W+e{{V}_{0}}\] ...(2) \[\therefore \] \[W=\left[ \frac{hc}{270}-\frac{hc}{350} \right]\times {{10}^{9}}\] \[W=\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}\times {{10}^{9}}\times (350-270)}{350\times 270\times 1.6\times {{10}^{-19}}}eV\]\[W=1.05eV\] Note: Energy of incident radiation should be more than the work function of the metal for photoelectrqhs to be emitted.
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