For intensity $I$ of a light of wavelength $5000 \, Å$, the photoelectron saturation current is $0.40 \, μA$ and the stopping potential is $1.36 \, V$. The work function of the metal is ........... $eV$.

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 following diagram,if $V_2 > V_1$,then:

Initially,a photon of wavelength $\lambda_1$ falls on a photocathode and emits an electron of maximum energy $E_1$. If the wavelength of the incident photon is changed to $\lambda_2$,the maximum energy of the electron emitted becomes $E_2$. Then the value of $hc$ ($h=$ Planck's constant,$c=$ velocity of light) is

In a photoelectric effect experiment,the threshold wavelength of incident light is $380 \, nm$. If the wavelength of the incident light is $260 \, nm$,the maximum kinetic energy of emitted electrons will be: .............. $eV$. Given $E \text{ (in } eV) = \frac{1237}{\lambda \text{ (in } nm)}$.

The threshold frequency of a photoelectric metal is $\nu_0$. If light of frequency $4\nu_0$ is incident on this metal,then the maximum kinetic energy of emitted electrons will be $.......h\nu_0$.