Light of two different frequencies whose photons have energies $1 \text{ eV}$ and $2.5 \text{ eV}$ respectively,successively illuminates a metal of work function $0.5 \text{ eV}$. The ratio of the maximum kinetic energies of the emitted electrons will be:

Photoelectric emission occurs only when the incident light has more than a certain minimum

When a metallic surface is illuminated with radiation of wavelength $\lambda$,the stopping potential is $V$. If the same surface is illuminated with radiation of wavelength $3\lambda$,the stopping potential is $\frac{V}{6}$. The threshold wavelength for the surface is:

The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength $6630 \; \mathring{A}$ is $0.42 \; V$. If the threshold frequency is $x \times 10^{13} \; s^{-1}$,where $x$ is ... (nearest integer).
(Given: speed of light $= 3 \times 10^{8} \; m/s$,Planck's constant $= 6.63 \times 10^{-34} \; J \cdot s$)

$A$ silver ball is suspended by a string in a vacuum chamber and ultraviolet light of wavelength $200 \ nm$ is incident on it. The maximum potential acquired by the ball,if the work function of silver is $4.7 \ eV$,will be .............. $V$.

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 $\,nm$)