Light of two different frequencies whose photons have energies $1.3 eV$ and $2.8 eV$ respectively,successfully illuminate a metallic surface whose work function is $0.8 eV$. The ratio of maximum speeds of emitted electrons will be

Two light waves of wavelengths $600 \,nm$ and $200 \,nm$ are incident on a metal surface. The maximum velocity of photoelectrons produced due to one wavelength is $\frac{1}{3}$ of the maximum velocity of the photoelectrons produced due to the other wavelength. The work function of the metal is:

If the surface of a metal is successively exposed to radiation of wavelengths $\lambda_1 = 350 \, nm$ and $\lambda_2 = 450 \, nm$,the maximum velocity of photoelectrons differs by a factor of $2$. The work function of this metal is:

The maximum kinetic energies of photoelectrons emitted are $K_1$ and $K_2$ when lights of wavelengths $\lambda_1$ and $\lambda_2$ are incident on a metallic surface. If $\lambda_1 = 3 \lambda_2$,then:

In a photoelectric experiment, the slope of the graph drawn between stopping potential $(V_s)$ along the $y$-axis and the frequency $(\nu)$ of incident radiation along the $x$-axis is (Planck's constant $h = 6.6 \times 10^{-34} \text{ Js}$)

Light of wavelength $\lambda$ strikes a photo-sensitive surface and electrons are ejected with kinetic energy $E$. If the kinetic energy is to be increased to $2E$,the wavelength must be changed to $\lambda'$ where