The threshold frequency for a photosensitive metal is $3.3 \times 10^{14} \text{ Hz}$. If light of frequency $8.20 \times 10^{14} \text{ Hz}$ is incident on this metal,the cut-off voltage for the photoelectron emission is nearly ............ $V$.

Energy of a photoelectron depends on which factors?

For the photoelectric effect with incident photon wavelength $\lambda$,the stopping potential is $V_0$. Identify the correct variation$(s)$ of $V_0$ with $\lambda$ and $1/\lambda$.

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

Two identical photocathodes receive light of frequencies $v_1$ and $v_2$. If the velocities of the photoelectrons (of mass $m$) coming out are $v_1$ and $v_2$ respectively,then:

When light of wavelength $300 \; nm$ falls on a photoelectric emitter,photoelectrons are liberated. For another emitter,light of $600 \; nm$ wavelength is sufficient for creating photoemission. What is the ratio of the work functions of the two emitters?