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  • Consider a metal exposed to light of wavelength 600 nm. The maximum energy of the electron doubles when light of wavelength 400 nm is used. Find the work function in eV.

Consider a metal exposed to light of wavelength 600 nm. The maximum energy of the electron doubles when light of wavelength 400 nm is used. Find the work function in eV.

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Let the maximum energies of emitted electrons be K1 and K2 when 600 nm and 400 nm visible light are used according to question

K_{2}=2K_{1}\\ K_{max}=hv-\phi=\frac{hc}{\lambda}-\phi\\ K_{1}=\frac{hc}{\lambda_{1}}-\phi\\ K_{2}=\frac{hc}{\lambda_{2}}-\phi=2K_{1}\\ \frac{hc}{\lambda_{2}}-\phi=2 \left [\frac{hc}{\lambda_{1}}-\phi \right ]=\frac{2hc}{\lambda_{1}}-2\phi\\ \phi=hc\left [ \frac{2}{\lambda_{1}}- \frac{1}{\lambda_{2}} \right ]\\ =1240\left [ \frac{2}{600}-\frac{1}{400} \right ]eV\\ =\frac{1240}{200}\left [ \frac{2}{3}-\frac{1}{2} \right ]=6.2\left ( \frac{4-3}{6} \right )\\ Work\; function\; \phi=\frac{6.2}{6}=1.03eV

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