The stopping potential for electrons emitted from a photosensitive surface illuminated by light of wavelength $491\, nm$ is $0.710\, V$. When the incident wavelength is changed to a new value,the stopping potential is $1.43\, V$. The new wavelength is ....... $nm.$

$A$ photosensitive surface has work function $\phi$. If a photon of energy $3 \phi$ falls on this surface, the electron comes out with a maximum velocity of $4 \times 10^6 \,m/s$. When the photon energy is increased to $7 \phi$, the maximum velocity of the photoelectron will be:

When light of a given wavelength is incident on a metallic surface,the minimum potential needed to stop the emitted photoelectrons is $6.0 \ V$. This potential drops to $0.6 \ V$ if another source with wavelength four times that of the first one and intensity half of the first one is used. What are the wavelength of the first source and the work function of the metal,respectively? $\left[\text{Take } hc = 1.24 \times 10^{-6} \ J \ m\right]$

Light of wavelength $\lambda$ strikes a photoelectric surface and electrons are ejected with energy $E$. If $E$ is to be increased to twice the original value,the wavelength changes to $\lambda_1$. Which of the following is correct?

Light from the Paschen series of a hydrogen atom is able to eject photoelectrons from a metal. Then the work function of the metal is:

In an experiment on the photoelectric effect,the frequency $\nu$ of the incident light is plotted against the stopping potential $V_0$. The work function of the photoelectric surface is given by ($e$ is the electronic charge):