The kinetic energy of the photoelectrons increases by $0.52 eV$ when the wavelength of incident light is changed from $500 \,nm$ to another wavelength. The new wavelength is approximately: (in $\,nm$)

In a photoelectric experiment,a monochromatic light is incident on the emitter plate $E$,as shown in the figure. When switch $S_1$ is closed and switch $S_2$ is open,the photoelectrons strike the collector plate $C$ with a maximum kinetic energy of $1 eV$. If switch $S_1$ is open and switch $S_2$ is closed and the frequency of the incident light is doubled,the photoelectrons strike the collector plate with a maximum kinetic energy of $20 eV$. The threshold wavelength of the emitter plate is (in $Å$)

The work function of a metallic surface is $5.01\ eV$. Photoelectrons are emitted when light of wavelength $2000\ \mathring{A}$ falls on it. The minimum potential difference required to stop the fastest photoelectrons is ................. $V$.

Which theory of light can explain the photoelectric effect?

Statement $-1$: When ultraviolet light is incident on a photocell, its stopping potential is $V_0$ and the maximum kinetic energy of the photoelectrons is $K_{max}$. When the ultraviolet light is replaced by $X$-rays, both $V_0$ and $K_{max}$ increase.
Statement $-2$: Photoelectrons are emitted with speeds ranging from zero to a maximum value because of the range of frequencies present in the incident light.

The graph of stopping potential $(V_{s})$ against frequency $(\nu)$ of incident radiation is plotted for two different metals '$P$' and '$Q$' as shown in the graph. If $\phi_{P}$ and $\phi_{Q}$ are the work functions of metals '$P$' and '$Q$' respectively,then which of the following is correct?