question_answer

Monochromatic radiation of wavelength 640.2nm \[(1nm={{10}^{-9}}m)\] from a neon lamp irradiates a photosensitive material made of calcium or tungsten. The stopping potential is measured to be 0.54V. The source is replaced by an iron source and its 427.2 nm line irradiates the same photocell. Predict the new stopping voltage.

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, \[{{V}_{0}}=\frac{2.42\times {{10}^{-19}}}{1.6\times {{10}^{-19}}}=1.51\,V\]                                     \[(\because e=1.6\times {{10}^{-19}}C)\]