Two photons with energies three times and nine times the work function of a metal surface are incident on it. What is the ratio of the maximum velocities of the photoelectrons emitted in the two cases?

When radiation of wavelength $\lambda$ is incident on a metallic surface,the stopping potential is $4.8 \ V$. If the same surface is illuminated with radiation of double the wavelength,then the stopping potential becomes $1.6 \ V$. Then the threshold wavelength for the surface is

The threshold frequency of a metal is $f_0$. When light of frequency $2f_0$ is incident on the metal plate, the maximum velocity of the photoelectrons is $v_1$. When the frequency of the incident radiation is increased to $5f_0$, the maximum velocity of the photoelectrons emitted is $v_2$. The ratio of $v_1$ to $v_2$ is

Initially,a photon of wavelength $\lambda_1$ falls on a photocathode and emits an electron of maximum energy $E_1$. If the wavelength of the incident photon is changed to $\lambda_2$,the maximum energy of the electron emitted becomes $E_2$. Then the value of $hc$ ($h=$ Planck's constant,$c=$ velocity of light) is

When a metallic surface is illuminated with a radiation of wavelength $\lambda$, the stopping potential is $V$. If the same surface is illuminated with radiation of wavelength $6\lambda$, the stopping potential is $V/12$. The threshold wavelength for the surface is

The electric field of a light wave is given as $\vec E = 10^{-3} \cos \left( \frac{2\pi x}{5 \times 10^{-7}} - 2\pi \times 6 \times 10^{14} t \right) \hat x \, N/C$. This light falls on a metal plate with a work function of $2 \, eV$. The stopping potential of the photoelectrons is ................ $V$.