In the experiment of $P.E.E.$,the $KE_{max}$ of an electron is $K_0$. If the frequency is increased by a factor of $n_1$,then the $KE_{max}$ becomes $n_2K_0$. Find the work function.

When ultraviolet rays are incident on a metal plate, the photoelectric effect does not occur. It occurs by the incidence of:

The threshold wavelength of a metal surface is $5 \times 10^{-10} \, m$. When it is illuminated by light of wavelength $2 \times 10^{-10} \, m$,the stopping potential is $V_0$. What will be the stopping potential if the wavelength of the incident light is doubled?

If the work function of a metal is $3 \ eV$,then the threshold wavelength will be ............ $\mathring{A}$.

Energy of the incident photons on the metal surface is initially $4W$ and then $6W$,where $W$ is the work function of that metal. The ratio of the maximum velocities of the emitted photoelectrons is:

The electric field component of an $EM$ radiation varies with time as $E = a(\cos \omega_{0} t + \sin \omega t \cos \omega_{0} t)$,where '$a$' is a constant,$\omega = 10^{15} \text{ s}^{-1}$,and $\omega_{0} = 5 \times 10^{15} \text{ s}^{-1}$. This radiation falls on a metal whose stopping potential is $2 \text{ V}$. Which of the following statement$(s)$ is/are true? $(h = 6.62 \times 10^{-34} \text{ J s})$