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 $1/\lambda$.

When radiations of wavelength $\lambda$ are incident on a metallic surface,the stopping potential required is $4.8 \ V$. If the same surface is illuminated with radiations of double the wavelength,the required stopping potential becomes $1.6 \ V$. The value of the threshold wavelength for the surface is:

The stopping potential for a source of wavelength $4000\, \mathring{A}$,when kept at a distance of $10\, \text{cm}$,is $1.5\, \text{V}$. If now the distance of the source is increased to $20\, \text{cm}$,then the stopping potential will be ............... $\text{V}$.

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?

For zero photoelectric current,the stopping potential is:

The electric field component of an $EM$ radiation varies with time as $E = a(\cos \omega_{0} t + \sin \omega t \cos \omega_{0} t)$, where '$a$' is a constant, $\omega = 10^{15} \text{ s}^{-1}$, and $\omega_{0} = 5 \times 10^{15} \text{ s}^{-1}$. This radiation falls on a metal whose stopping potential is $2 \text{ V}$. Which of the following statement$(s)$ is/are true? $(h = 6.62 \times 10^{-34} \text{ J s})$