The graph between the maximum speed $v_{\text{max}}$ of a photoelectron and frequency $(\nu)$ of the incident radiation,in the photoelectric effect,is correctly represented by:

$A$ photon released by the transition of an electron from the second excited state to the ground state of a Hydrogen atom is incident on the surface of a metal with a work function of $3.1 \ eV$. The de Broglie wavelength of the most energetic electron emitted from that metal surface is nearly:

When light of a given wavelength is incident on a metallic surface,the minimum potential needed to stop the emitted photoelectrons is $6.0 \ V$. This potential drops to $0.6 \ V$ if another source with wavelength four times that of the first one and intensity half of the first one is used. What are the wavelength of the first source and the work function of the metal,respectively? $\left[\text{Take } hc = 1.24 \times 10^{-6} \ J \ m\right]$

The threshold wavelength for photoelectric effect from a metal is $200 \ nm$. The maximum kinetic energy of the emitted photoelectrons when radiation of wavelength $100 \ nm$ is incident on the metal is .......... $eV$.

When a metallic surface is illuminated with radiation of wavelength $\lambda$,the stopping potential is $V$. If the same surface is illuminated with radiation of wavelength $2\lambda$,the stopping potential is $\frac{V}{4}$. The threshold wavelength for the metallic surface is:

When a metallic surface is illuminated with light of wavelength $\lambda$,the stopping potential is $x \ V$. When the same surface is illuminated by light of wavelength $2 \lambda$,the stopping potential is $\frac{x}{3} \ V$. The threshold wavelength for the metallic surface is ..........