$A$ monochromatic neon lamp with a wavelength of $670.5\, nm$ illuminates a photo-sensitive material which has a stopping voltage of $0.48\, V$. What will be the stopping voltage if the source light is changed with another source of wavelength $474.6\, nm$? (In $V$)

The figure showing the correct relationship between the stopping potential $V_0$ and the frequency $\nu$ of light for potassium and tungsten is

The graph shows the variation of stopping potential $V_o$ with the frequency $\nu$ of the incident radiation for three photosensitive metals $X_1, X_2$ and $X_3$. Which metal will emit photoelectrons with greater kinetic energy,for the same wavelength of incident radiation?

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 $\,nm$)

The work function of caesium metal is $2.14 \; eV$. When light of frequency $6 \times 10^{14} \; Hz$ is incident on the metal surface,photoemission of electrons occurs. What is the
$(a)$ maximum kinetic energy of the emitted electrons,
$(b)$ stopping potential,and
$(c)$ maximum speed of the emitted photoelectrons?

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: