The frequency of incident light falling on a photosensitive material is doubled. The kinetic energy $(K.E.)$ of the emitted photoelectrons will be:

$A$ metal surface is illuminated by light of wavelength $400 \ nm$. The kinetic energy of the emitted photoelectrons is found to be $1.68 \ eV$. The work function of the metal is ......... $eV$. $(hc = 1240 \ eV \cdot nm)$

The wavelengths of light incident on a photocell are $400 \ nm$ and $250 \ nm$. The velocities of the emitted photoelectrons are $v$ and $2v$ respectively. What is the work function of the metal?

When the light of frequency $2v_0$ (where $v_0$ is threshold frequency) is incident on a metal plate,the maximum velocity of electrons emitted is $v_1$. When the frequency of the incident radiation is increased to $5v_0$,the maximum velocity of electrons emitted from the same plate is $v_2$. The ratio of $v_1$ to $v_2$ is

When monochromatic light falls on a photo-sensitive metal,an electron is emitted with maximum velocity $1.6 \times 10^6 \ m/s$. Find the stopping potential.
[charge of electron $= 1.6 \times 10^{-19} \ C$,mass of electron $= 9 \times 10^{-31} \ kg$] (in $V$)

All electrons ejected from a metallic surface by incident light of wavelength $400 \,nm$ travelled $1 \,m$ in the direction of a uniform electric field of $2 \,N/C$ and came to rest. The work function of the surface is (in $\,eV$)