When a metallic surface is illuminated with a radiation of wavelength $\lambda$, the stopping potential is $V$. If the same surface is illuminated with radiation of wavelength $6\lambda$, the stopping potential is $V/12$. The threshold wavelength for the surface is

When two monochromatic lights of frequency $v$ and $\frac{v}{2}$ are incident on a photoelectric metal,their stopping potentials are $\frac{V_{s}}{2}$ and $V_{s}$ respectively. The threshold frequency for this metal is:

When radiation of wavelength $\lambda$ is incident on a metal,the stopping potential of photoelectrons is $4.8 \ V$. When radiation of wavelength $2\lambda$ is incident on the same metal,the stopping potential is $1.6 \ V$. What is the threshold wavelength of the metal?

When photons are incident on a photosensitive material of work function $1.5 \text{ eV}$,the maximum velocity of the emitted photoelectrons is $8 \times 10^5 \text{ m/s}$. The stopping potential of the photoelectrons is (Mass of the electron $= 9 \times 10^{-31} \text{ kg}$ and charge of the electron $= 1.6 \times 10^{-19} \text{ C}$) (in $V$)

An electron in a hydrogen atom jumps from the second Bohr orbit to the ground state,and the energy difference of the two states is radiated in the form of photons. These photons are then allowed to fall on a metal surface having a work function equal to $4.2 \ eV$. Calculate the stopping potential. [Energy of electron in $n^{\text{th}}$ orbit $= -\frac{13.6}{n^2} \ eV$] (in $V$)

$A$ metal surface is illuminated by light of given intensity and frequency to cause photoemission. If the intensity of illumination is reduced to one fourth of its original value,then the maximum $KE$ of the emitted photoelectrons would be