The surface of a metal is first illuminated with a light of wavelength $300 \ nm$ and later illuminated by another light of wavelength $500 \ nm$. It is observed that the ratio of maximum velocities of photoelectrons in the two cases is $3$. The work function of the metal is close to: (in $eV$)

An electron moving with velocity $1.6 \times 10^7 \ m/s$ has a wavelength of $0.4 \ \mathring{A}$. The required accelerating voltage for the electron motion is (charge on electron $= 1.6 \times 10^{-19} \ C$,mass of electron $= 9 \times 10^{-31} \ kg$).

Given that a photon of light of wavelength $10,000\;\mathring A$ has an energy equal to $1.23\; eV$. When light of wavelength $5000\;\mathring A$ and intensity $I_0$ falls on a photoelectric cell,the saturation current is $0.40 \times 10^{-6}\; A$ and the stopping potential is $1.36\; V$; then the work function is ..... $eV$.

The kinetic energy of the photoelectrons increases by $0.52 eV$ when the wavelength of incident light is changed from $500 \,nm$ to another wavelength. The new wavelength is approximately: (in $\,nm$)

Radiation of two photon energies,twice and five times the work function of a metal,are incident successively on the metal surface. The ratio of the maximum velocity of photoelectrons emitted in the two cases is:

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: