Answer:
1.51 V Here, for neon lamp, \[\lambda =640.2\,\,nm=640.2\times {{10}^{-9}}m\] \[{{V}_{0}}=0.54\,V\] We know that, \[e{{V}_{0}}=\frac{hc}{\lambda }-{{\phi }_{0}}\] \[\therefore \] Work function, \[{{\phi }_{0}}=\frac{hc}{\lambda }-e{{V}_{0}}\] \[=\frac{6.63\times {{10}^{-34}}\times 3\times {{10}^{8}}}{640.2\times {{10}^{-9}}}-1.6\times {{10}^{-9}}\times 0.54\] \[=(3.1\times {{10}^{-19}}-0.864\times {{10}^{-19}})\,\text{J}\] \[=2.243\times {{10}^{-19}}\text{J=}\frac{2.243\times {{10}^{-19}}}{1.6\times {{10}^{-19}}}eV\simeq 1.4eV\] For iron source \[\lambda =427.2\,nm=427.2\times {{10}^{-9}}m\] \[\therefore \] \[e{{V}_{0}}=\frac{hc}{\lambda }={{\phi }_{0}}\] \[=\frac{6.63\times {{10}^{-34}}\times 3\times {{10}^{8}}}{427.2\times {{10}^{-9}}}-2.236\times {{10}^{-19}}\] \[=2.42\times {{10}^{-19}}\text{J}\] \[\therefore \] Stopping potential, \[{{V}_{0}}=\frac{2.42\times {{10}^{-19}}}{1.6\times {{10}^{-19}}}=1.51\,V\] \[(\because e=1.6\times {{10}^{-19}}C)\]
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