$A$ photon of energy $5.5 \ eV$ strikes a surface that emits photoelectrons with a maximum kinetic energy of $4.0 \ eV$. The stopping potential for these electrons is ............ $V$.

An electron and a proton are separated by a large distance. The electron starts approaching the proton with an initial kinetic energy of $3 \, eV$. The proton captures the electron and forms a hydrogen atom in the second excited state. The resulting photon is incident on a photosensitive metal with a threshold wavelength of $4000 \, Å$. What is the maximum kinetic energy of the emitted photoelectron? (In $eV$)

The work function of metals is in the range of $2 eV$ to $5 eV$. Find which of the following wavelengths of light cannot be used for the photoelectric effect (in $nm$)? (Consider,Planck constant $= 4 \times 10^{-15} eVs$,velocity of light $= 3 \times 10^{8} m/s$)

Initially,a photon of wavelength $\lambda_1$ falls on a photocathode and emits an electron of maximum energy $E_1$. If the wavelength of the incident photon is changed to $\lambda_2$,the maximum energy of the electron emitted becomes $E_2$. Then the value of $hc$ ($h=$ Planck's constant,$c=$ velocity of light) is

Which of the following statements is correct in the case of the photoelectric effect?

$A$ photosensitive surface has work function $\phi$. If a photon of energy $3 \phi$ falls on this surface, the electron comes out with a maximum velocity of $4 \times 10^6 \,m/s$. When the photon energy is increased to $7 \phi$, the maximum velocity of the photoelectron will be: