The work function of a metal surface is $1.6 \times 10^{-19} \text{ J}$. When light of wavelength $6400 \ \mathring{A}$ is incident on this surface,the maximum kinetic energy of the emitted photoelectrons is ....... $\text{J}$.

For a photosensitive material, the work function is $W_0$ and the stopping potential is $V$. What is the wavelength of the incident radiation? ($h$ = Planck's constant, $c$ = velocity of light, $e$ = electronic charge)

$A$ photo-emissive substance is illuminated with a radiation of wavelength $\lambda_i$ so that it releases electrons with de-Broglie wavelength $\lambda_c$. The longest wavelength of radiation that can emit photoelectrons is $\lambda_0$. The expression for the de-Broglie wavelength is given by ($m$: mass of the electron,$h$: Planck's constant,and $c$: speed of light).

The maximum kinetic energy of a photoelectron is $E$ when the wavelength of incident radiation is $\lambda$. If the wavelength of the incident radiation is reduced to $\frac{\lambda}{3}$,the maximum kinetic energy becomes $4E$. The work function of the metal is:

Maximum velocity of the photoelectron emitted by a metal is $1.8 \times 10^{6} \ m/s$. Take the value of specific charge of the electron as $1.8 \times 10^{11} \ C/kg$. Then the stopping potential in volt is

If in a photoelectric experiment,the wavelength of incident radiation is reduced from $6000 \, \mathring{A}$ to $4000 \, \mathring{A}$,then: