The threshold wavelength of tungsten is $2300 \, \mathring{A}$. If ultraviolet light of wavelength $1800 \, \mathring{A}$ is incident on it,then the maximum kinetic energy of photoelectrons would be about ............ $eV$.

$A$ beam of light of wavelength $400\,nm$ and power $1.55\,mW$ is directed at the cathode of a photoelectric cell. If only $10\%$ of the incident photons effectively produce photoelectrons,then find the current due to these electrons in $\mu A$. (Given: $hc = 1240\,eV\cdot nm$,$e = 1.6 \times 10^{-19}\,C$)

The surface of a metal is illuminated with light of $400 \ nm$. The kinetic energy of the ejected photoelectrons was found to be $1.68 \ eV$. The work function of the metal is ............ $eV$ $(hc = 1240 \ eV \ nm)$.

Statement $1$: When ultraviolet light is incident on a photocell, its stopping potential is $V_0$ and the maximum kinetic energy of photoelectrons is $K_{max}$. When $X$-rays are used instead of ultraviolet light, both $V_0$ and $K_{max}$ increase.
Statement $2$: Photoelectrons are emitted with a range of speeds from $0$ to a maximum value because the incident light contains a range of frequencies.

In the experiment of $P.E.E.$ (Photoelectric Effect), the saturation current is $5\,mA$ and the stopping potential is $10\,V$. If the intensity and frequency of the incident light are both doubled, then what will be the new saturation current $(i_s)$ and stopping potential $(V_s)$?

In a photoelectric emission experiment,the stopping potential for a given metal is $V$ volt,when radiation of wavelength $\lambda$ is used. If radiation of wavelength $2 \lambda$ is used with the same metal,then the stopping potential (in volt) will be. [Given: $c = \text{velocity of light}$,$e = \text{charge on electron}$,$h = \text{Planck's constant}$]