When a photon of energy $hv$ is incident on an aluminum plate (work function $E_0$),photoelectrons with maximum kinetic energy $K$ are emitted. What will be the maximum kinetic energy of the emitted photoelectrons when a photon of energy $2hv$ is incident on the same aluminum plate?

$A$ light whose frequency is equal to $6 \times 10^{14} \, Hz$ is incident on a metal whose work function is $2 \, eV$. $[h = 6.63 \times 10^{-34} \, Js, 1 \, eV = 1.6 \times 10^{-19} \, J]$. The maximum kinetic energy of the emitted electrons will be ............ $eV$. (in $.49$)

$A$ photoelectric cell is connected to a source of variable potential difference. The photoelectric current is plotted against the applied potential difference. The graph with the broken line represents the current for a given frequency and intensity of the incident radiation. If the frequency is increased and the intensity is reduced, which curve now represents the situation?

When a metal is irradiated by light having wavelength $\lambda$ $(\lambda < \lambda_0)$,all the photoelectrons emitted are bent in a circle of radius $r$ by a magnetic field of flux density $B_0$. Find $\frac{1}{\lambda_0}$,where $\lambda_0$ is the threshold wavelength.

Light of wavelength $\lambda$ which is less than threshold wavelength is incident on a photosensitive material. If incident wavelength is decreased so that emitted photoelectrons are moving with the same velocity,then stopping potential will:

When monochromatic light falls on a photo-sensitive metal,an electron is emitted with maximum velocity $1.6 \times 10^6 \ m/s$. Find the stopping potential.
[charge of electron $= 1.6 \times 10^{-19} \ C$,mass of electron $= 9 \times 10^{-31} \ kg$] (in $V$)