The work function of nickel is $5 \text{ eV}$. When light of wavelength $2000 \text{ Å}$ falls on it,it emits photoelectrons. The potential difference necessary to stop the fastest emitted electrons is (given $h = 6.67 \times 10^{-34} \text{ J-s}$): (in $\text{ V}$)

In the experiment of the photoelectric effect, if the frequency of incident light is $v_1$, the maximum kinetic energy $(K.E.)$ of photoelectrons is $K_0$. If the frequency of light is $v_2$, the maximum $K.E.$ of photoelectrons is $2K_0$. Which of the following relations is correct?

In a photoelectric experiment, the slope of the graph drawn between stopping potential $(V_s)$ along the $y$-axis and the frequency $(\nu)$ of incident radiation along the $x$-axis is (Planck's constant $h = 6.6 \times 10^{-34} \text{ Js}$)

According to Einstein's photoelectric equation,the graph of the kinetic energy of photoelectrons emitted from a metal versus the frequency of incident radiation is a straight line whose slope:

If the threshold wavelength of light for photoelectric emission to take place from a metal surface is $6250 \ \text{Å}$, then the work function of the metal is (Planck's constant $= 6.6 \times 10^{-34} \ \text{Js}$) (in $\text{eV}$)

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 $Å$)