In a photoelectric experiment,light of energy $2.48 eV$ irradiates a photosensitive material. The stopping potential was measured to be $0.5 V$. The work function of the photosensitive material is: (in $eV$)

Light of wavelength $\lambda$ falls on a metal having work function $\frac{hc}{\lambda_0}$. Photoelectric effect will take place only if ($\lambda_0$ is the threshold wavelength).

$A$ photoelectric material having work-function $\phi_0$ is illuminated with light of wavelength $\lambda$ (where $\lambda < \frac{hc}{\phi_0}$). The fastest photoelectron has a de Broglie wavelength $\lambda_d$. $A$ change in wavelength of the incident light by $\Delta \lambda$ results in a change $\Delta \lambda_d$ in $\lambda_d$. Then the ratio $\frac{\Delta \lambda_d}{\Delta \lambda}$ is proportional to:

The frequency of incident light falling on a photosensitive material is doubled. The kinetic energy $(K.E.)$ of the emitted photoelectrons will be:

When the wavelength of incident radiation on a metal surface is reduced from $\lambda_1$ to $\lambda_2$,the kinetic energy of the emitted photoelectrons is tripled. Find the work function of the metal. [$h =$ Planck's constant,$c =$ velocity of light]

The work function of Cesium is $2.27 eV$. The cut-off voltage which stops the emission of electrons from a cesium cathode irradiated with light of $600 nm$ wavelength is