The maximum velocity of the photoelectron emitted by the metal surface is $V$. The charge and mass of the photoelectron are denoted by $e$ and $m$ respectively. The stopping potential in volts is:

Light of frequency $v$ falls on a material of threshold frequency $v_0$. Maximum kinetic energy of emitted electron is proportional to

The ratio of work functions of two metals $A$ and $B$ is $1:2$. If radiations of frequencies $f$ and $2f$ are incident on the surfaces of $A$ and $B$ respectively,the ratio of the maximum kinetic energies of the emitted photoelectrons will be: (Given $f >$ threshold frequency of $A$ and $2f >$ threshold frequency of $B$)

Cut-off potentials for a metal in the photoelectric effect for light of wavelengths $\lambda_1$,$\lambda_2$,and $\lambda_3$ are found to be $V_1$,$V_2$,and $V_3$ volts. If $V_1$,$V_2$,and $V_3$ are in Arithmetic Progression,then $\lambda_1$,$\lambda_2$,and $\lambda_3$ will be in:

Light of wavelength $\lambda_{\text{ph}}$ falls on a cathode plate inside a vacuum tube as shown in the figure. The work function of the cathode surface is $\phi$ and the anode is a wire mesh of conducting material kept at a distance $d$ from the cathode. $A$ potential difference $V$ is maintained between the electrodes. If the minimum de Broglie wavelength of the electrons passing through the anode is $\lambda_e$,which of the following statement$(s)$ is(are) true?

When the wavelength of an incident photon is decreased, then: