Radiation of wavelength $332 \ nm$ is incident on a metal surface having a work function of $1.07 \ eV$. The stopping potential required to stop the emission of photoelectrons from the metal surface is ............ $V$. $(h = 6.6 \times 10^{-34} \ J s, c = 3 \times 10^8 \ m/s, 1 \ eV = 1.6 \times 10^{-19} \ J)$

$A$ beam of light of wavelength $\lambda$ and with illumination $L$ falls on a clean surface of sodium. If $N$ photoelectrons are emitted each with kinetic energy $E$,then

In the experiment of photoelectric effect, $K_{max}$ for a photoelectron is $K_0$. If the frequency of light is doubled, then $K_{max}$ is $3K_0$. What would be the value of $K_{max}$ if the frequency of light is tripled? (in terms of $K_0$)

When a photon of energy $6 \ eV$ is incident on a metal surface with a work function of $2.1 \ eV$,what is the stopping potential of the emitted electrons in $V$?

Photoelectric emission is observed from a metallic surface for frequencies $v_1$ and $v_2$ of the incident light rays $(v_1 > v_2)$. If the maximum values of kinetic energy of the photoelectrons emitted in the two cases are in the ratio of $1:k$,then the threshold frequency of the metallic surface is

If the electron in a hydrogen atom jumps from the third Bohr orbit to the ground state directly and the difference between the energies of the two states is radiated in the form of photons. If the work function of the material is $4.1 \text{ eV}$, then the stopping potential is nearly:
$\left[\text{Energy of electron in } n^{\text{th}} \text{ orbit} = \frac{-13.6}{n^2} \text{ eV}\right]$ (in $\text{ V}$)