The kinetic energy of an emitted electron is $E$ when the light incident on the metal has wavelength $\lambda$. To double the kinetic energy,the incident light must have a wavelength of:

The slope of the graph between the frequency of incident light and the stopping potential for a given surface is:

Radiation $p$ with wavelength $6561 \; \mathring{A}$ falls on a metal surface to produce photoelectrons. The electrons are made to enter a uniform magnetic field of $3 \times 10^{-4} \; T$. If the radius of the largest circular path followed by the electrons is $10 \; mm$,the work function of the metal is close to ............... $eV$.

Light described at a place by the equation $\varepsilon = (100 \text{ V/m}) [\sin(5 \times 10^{15} \text{ s}^{-1})t + \sin(8 \times 10^{15} \text{ s}^{-1})t]$ falls on a metal surface having a work function of $2 \text{ eV}$. Calculate the maximum kinetic energy of the photoelectrons in $\text{eV}$.

Photoelectric effect is observed on a surface for frequencies $5.5 \times 10^8 \text{ MHz}$ and $4.5 \times 10^8 \text{ MHz}$. If the ratio of the maximum kinetic energy of the emitted photoelectrons is $1:5$,then the threshold frequency for the metal surface is:

Light is incident on a metallic plate having a work function of $ 110 \times 10^{-20} \ J $. If the produced photoelectrons have zero kinetic energy, then the angular frequency of the incident light is . . . . . . $ rad/s $. $( h = 6.63 \times 10^{-34} \ J \cdot s )$