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 ............... $\text{V}$.

Photons of energy $2.4 \text{ eV}$ and wavelength $\lambda$ fall on a metal plate and release photoelectrons with a maximum velocity $v$. By decreasing $\lambda$ by $50 \%$, the maximum velocity of photoelectrons becomes $3 v$. The work function of the material of the metal plate is (in $\text{ eV}$)

The stopping potential doubles when the frequency of the incident light changes from $v$ to $\frac{3v}{2}$. Then the work function of the metal must be

The beam of light has three wavelengths $4144 \,\mathring A$,$4972 \,\mathring A$,and $6216 \,\mathring A$ with a total intensity of $3.6 \times 10^{-5} \, W/m^2$ equally distributed amongst the three wavelengths. The beam falls normally on an area of $1 \, cm^2$ of a clean metallic surface with a work function of $2.3 \, eV$. Assume that there is no loss of light by reflection and that each energetically capable photon ejects one electron. Calculate the number of photoelectrons liberated in $2 \, s$.

The photoelectric threshold wavelength for a metal surface is $6600 \, \mathring{A}$. The work function for this is:

The threshold frequency of a photosensitive material is equal to the frequency of the $H_{\alpha}$ line of hydrogen. If a photon whose frequency is equal to the frequency of the $H_{\beta}$ line of hydrogen is incident on this photosensitive material,the maximum kinetic energy of the emitted photoelectrons is ($R$ = Rydberg's constant,$h$ = Planck's constant,and $c$ = speed of light in vacuum).