The photoelectric threshold frequency of a metal is $v$. When light of frequency $4v$ is incident on the metal,the maximum kinetic energy of the emitted photoelectrons is

In a photocell circuit,the stopping potential $V_0$ is a measure of the maximum kinetic energy of the photoelectrons. The following graph shows experimentally measured values of stopping potential versus frequency $\nu$ of incident light. The values of Planck's constant and the work function as determined from the graph are (taking the magnitude of electronic charge to be $e = 1.6 \times 10^{-19} \, C$):

$A$ light of wavelength $310 \,nm$ is used in a photoelectric experiment. The metal electrode of work function $2.5 \,eV$ is used in the experiment. The stopping potential for the photoelectrons will be (assume $hc = 1240 \,eV-nm$): (in $V$)

$A$ 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:

If the frequency of light falling on a photosensitive material doubles, which of the following is true?

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?