In a photoelectric effect experiment,the threshold wavelength of incident light is $380 \, nm$. If the wavelength of the incident light is $260 \, nm$,the maximum kinetic energy of emitted electrons will be: .............. $eV$. Given $E \text{ (in } eV) = \frac{1237}{\lambda \text{ (in } nm)}$.

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

$A$ photoelectric surface is illuminated successively by monochromatic light of wavelength $\lambda$ and $\lambda /2$. If the maximum kinetic energy of the emitted photoelectrons in the second case is $3$ times that in the first case,the work function of the surface of the material is
$(h =$ Planck's constant,$c =$ speed of light $)$

In the photoelectric effect, the stopping potential depends on:

The maximum wavelength of light which causes photoelectric emission from a photosensitive metal surface is $\lambda_0$. Two light beams of wavelengths $\frac{\lambda_0}{3}$ and $\frac{\lambda_0}{9}$ are incident on the metal surface. The ratio of the maximum velocities of the emitted photoelectrons is

If light of wavelength $\lambda_1$ is allowed to fall on a metal,then the kinetic energy of the photoelectrons emitted is $E_1$. If the wavelength of light changes to $\lambda_2$,then the kinetic energy of the electrons changes to $E_2$. Then the work function of the metal is: