The light of two different frequencies whose photons have energies $3.8 \, eV$ and $1.4 \, eV$ respectively,illuminate a metallic surface whose work function is $0.6 \, eV$ successively. The ratio of maximum speeds of emitted electrons for the two frequencies respectively will be

Two incident radiations having energies two times and ten times of the work function of a metal surface,produce photoelectric effect. The ratio of maximum velocities of emitted photoelectrons respectively is

The surface of a certain metal is first illuminated with light of wavelength $\lambda_1 = 350 \ nm$ and then with light of wavelength $\lambda_2 = 540 \ nm$. It is found that the maximum speed of the photoelectrons in the two cases differs by a factor of $2$. The work function of the metal (in $eV$) is close to: (Energy of photon $= \frac{1240}{\lambda \text{ (in } nm)} \ eV$)

Work function of lithium and copper are respectively $2.3 \ eV$ and $4.0 \ eV$. Which one of the metal will be useful for the photoelectric cell working with visible light? $(h = 6.6 \times 10^{-34} \ J \cdot s, c = 3 \times 10^8 \ m/s)$

In the experiment of the photoelectric effect,the stopping potential for a given metal is $V_{0}$ volts when radiation of wavelength $\lambda_{0}$ is used. If radiation of wavelength $2\lambda_{0}$ is used for the same metal,then the stopping potential (in volts) will be: [$e=$ charge on electron,$c=$ speed of light,$h=$ Planck's constant.]

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: