A) \[4\times {{10}^{-59}}Vs\]
B) \[0.25\times {{10}^{55}}Vs\]
C) \[4\times {{10}^{-15}}Vs\]
D) \[4\times {{10}^{-8}}Vs\]
Correct Answer: C
Solution :
From Einsteins photoelectric equation, the maximum kinetic energy\[({{E}_{k}})\]of electrons given by \[{{E}_{k}}=hv-W\] where W is work function of metal, v the frequency. Also, \[{{E}_{k}}=eV\]and\[v=\frac{c}{\lambda }\] \[\therefore \] \[eV=\frac{hc}{\lambda }-W\] \[0.5e=\frac{hc}{6\times {{10}^{-7}}}-W\] \[\Rightarrow \] \[0.5=\frac{h}{e}.\frac{c}{6\times {{10}^{-7}}}-\frac{W}{e}\] ?. (i) and \[1.5=\frac{h}{e}\frac{c}{4\times {{10}^{-7}}}-\frac{W}{e}\] ...(ii) Subtracting Eq. (ii) from Eq. (i), we get \[1=\frac{h}{e}.\frac{c}{{{10}^{-7}}}\left[ \frac{1}{4}-\frac{1}{6} \right]\] \[\Rightarrow \] \[\frac{h}{e}=\frac{12\times {{10}^{-7}}}{3\times {{10}^{8}}}=4\times {{10}^{-15}}Vs\]
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