When radiation of wavelength $\lambda$ is used to illuminate a metallic surface,the stopping potential is $V.$ When the same surface is illuminated with radiation of wavelength $3 \lambda,$ the stopping potential is $\frac{V}{4}.$ If the threshold wavelength for the metallic surface is $n \lambda,$ then the value of $n$ will be......

When light of wavelength $\lambda$ is incident on a photosensitive surface,the stopping potential is $V$. When light of wavelength $3 \lambda$ is incident on the same surface,the stopping potential is $\frac{V}{6}$. Then the threshold wavelength for the surface is:

Two incident radiations having energies two times and ten times of the work function of a metal surface,produce photoelectric effect. The ratio of maximum velocities of emitted photoelectrons respectively is

$A$ Hydrogen-like atom has atomic number $Z$. Photons emitted in the electronic transitions from level $n=4$ to level $n=3$ in these atoms are used to perform a photoelectric effect experiment on a target metal. The maximum kinetic energy of the photoelectrons generated is $1.95 \ eV$. If the photoelectric threshold wavelength for the target metal is $310 \ nm$,the value of $Z$ is:

$A$ 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:

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: