When photons of energy $5 eV$ are incident on a metal surface with a work function of $4.36 eV$,photoelectrons are emitted. What is the maximum momentum of the emitted photoelectrons?

When a light of wavelength $4900 Å$ falls on a photosensitive metal, a negative $2 \,V$ potential is required to stop the emitted electrons. Then, the work-function of the material is nearly (given charge on electron $= 1.602 \times 10^{-19} C$ and Planck's constant $= 6.625 \times 10^{-34} Js$) (in $eV$)

When the light of frequency $2v_0$ (where $v_0$ is threshold frequency) is incident on a metal plate,the maximum velocity of electrons emitted is $v_1$. When the frequency of the incident radiation is increased to $5v_0$,the maximum velocity of electrons emitted from the same plate is $v_2$. The ratio of $v_1$ to $v_2$ is

When radiation of wavelength $400 \ nm$ is incident on a metal surface,the kinetic energy of the emitted photoelectrons is $1.68 \ eV$. The work function of the metal is .......... $eV$.

In a photoelectric effect experiment,the cathode metal is exposed to light of wavelength $600 \ nm$. The maximum kinetic energy of the ejected electron doubles when light of wavelength $400 \ nm$ is used. The work function of the cathode metal is approximately: [Use $h=6.63 \times 10^{-34} \ J-s, c=3 \times 10^8 \ m/s$] (in $eV$)

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 $\text{ V}$)