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(D) According to Einstein's photoelectric equation: $E = \phi + K_{max}$.Here,$E$ is the energy of the incident photon,$\phi$ is the work function,and

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$1.41$

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(D) According to Einstein's photoelectric equation: $E = \phi + K_{max}$.Here,$E$ is the energy of the incident photon,$\phi$ is the work function,and $K_{max}$ is the maximum kinetic energy of the photoelectrons.The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.Given $hc = 1240 \, eV-nm$ and $\lambda = 400 \, nm$,we have $E = \frac{1240}{400} \, eV = 3.1 \, eV$.Substituting the values into the photoelectric equation: $3.1 \, eV = \phi + 1.68 \, eV$.Therefore,the work function $\phi = 3.1 \, eV - 1.68 \, eV = 1.42 \, eV$.

The surface of a metal is illuminated with light of wavelength $400 \, nm$. The kinetic energy of the ejected photoelectrons is found to be $1.68 \, eV$. The work function of the metal is .................. $eV$ $(hc = 1240 \, eV-nm)$.

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