In an experiment of photoelectric effect,the stopping potential was measured to be $V_{1}$ and $V_{2}$ with incident light of wavelength $\lambda$ and $\frac{\lambda}{2}$ respectively. The relation between $V_{1}$ and $V_{2}$ is:

Monochromatic light of wavelength $\lambda = 4770 \ \mathring{A}$ is incident separately on the surface of four different metals $A, B, C$ and $D$. The work functions of $A, B, C$ and $D$ are $4.2 \ \text{eV}, 3.7 \ \text{eV}, 3.2 \ \text{eV}$ and $2.3 \ \text{eV}$,respectively. From which of these metals will electrons be emitted?

Which one of the following four graphs showing lines $P, Q, R$ and $S$ between maximum kinetic energy $(E)$ and intensity of incident light $(I)$ is correct?

The stopping potential for photoelectrons from a metal surface is $V_{1}$ when monochromatic light of frequency $v_{1}$ is incident on it. The stopping potential becomes $V_{2}$ when monochromatic light of another frequency is incident on the same metal surface. If $h$ is the Planck's constant and $e$ is the charge of an electron,then the frequency of light in the second case is

In a photoelectric experiment,the stopping potential was measured to be $V_1$ and $V_2$ volts with incident light of wavelength $\lambda$ and $\frac{\lambda}{2}$ respectively. The value of $V_2$ is: [where $\phi=$ work function,$e=$ electronic charge]

Light of frequency $4\nu_0$ is incident on a metal surface with threshold frequency $\nu_0$. The maximum kinetic energy of the emitted photoelectrons is: