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 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:

When a photosensitive material is illuminated by photons of energy $3.1 \ eV$,the stopping potential of the photoelectrons is $1.7 \ V$. When the same photosensitive material is illuminated by photons of energy $2.5 \ eV$,the stopping potential of the photoelectrons is: (in $V$)

When light falls on a metal surface,the maximum kinetic energy of the emitted photo-electrons depends upon

When light of frequency $v_{1}$ is incident on a metal with work function $W$ (where $h v_{1} > W$), then the photocurrent falls to zero at a stopping potential of $V_{1}$. If the frequency of light is increased to $v_{2}$, the stopping potential changes to $V_{2}$. Therefore, the charge of an electron $e$ is given by:

The threshold wavelength for sodium is $6 \times 10^{-7} \, m$. Photoemission occurs for light of wavelength $\lambda$ if:

The electron in the hydrogen atom jumps from excited state $(n = 3)$ to its ground state $(n = 1)$ and the photons thus emitted irradiate a photosensitive material. If the work function of the material is $5.1 \, eV$,the stopping potential is estimated to be (the energy of the electron in $n^{th}$ state $E_n = \frac{-13.6}{n^2} \, eV$) (in $, V$)