The work function of a photosensitive metal surface is $1.1 \ eV$. Two light beams of energies $1.5 \ eV$ and $2 \ eV$ are incident on the metal surface. The ratio of the maximum velocities of the emitted photoelectrons is

Assertion : Photosensitivity of a metal is high if its work function is small.
Reason : Work function $= hf_0$ where $f_0$ is the threshold frequency.

When a photosensitive metal surface is illuminated with radiation of wavelength $\lambda_1$,the stopping potential is $V_1$. If the same surface is illuminated with radiation of wavelength $3\lambda_1$,the stopping potential is $\frac{V_1}{6}$. The threshold wavelength for the photosensitive metal surface is:

When a point source of light is at a distance of $50 \, cm$ from a photoelectric cell,the cut-off voltage is found to be $V_0$. If the same source is placed at a distance of $1 \, m$ from the cell,then the cut-off voltage will be

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 $, V$)

The maximum velocity of an electron emitted by light of wavelength $\lambda$ incident on the surface of a metal of work function $\phi$ is [$h=$ Planck's constant,$m=$ mass of electron and $c=$ speed of light]