All electrons ejected from a metallic surface by incident light of wavelength $400 \,nm$ travelled $1 \,m$ in the direction of a uniform electric field of $2 \,N/C$ and came to rest. The work function of the surface is (in $\,eV$)

In the photoelectric effect, an electromagnetic wave is incident on a metal surface and electrons are ejected from the surface. If the work function of the metal is $2.14 \text{ eV}$ and the stopping potential is $2 \text{ V}$, what is the wavelength of the electromagnetic wave (in $\text{ nm}$)? (Given $hc = 1242 \text{ eV nm}$, where $h$ is Planck's constant and $c$ is the speed of light in vacuum.)

Light of wavelength $\lambda$ strikes a photoelectric surface and electrons are ejected with kinetic energy $K$. If $K$ is to be increased to twice its original value,the wavelength must be changed to $\lambda'$ such that :-

The photoelectric threshold wavelength for a certain metal surface is $3300 \,\mathring{A}$. What is the maximum kinetic energy of the photoelectrons emitted if radiations of wavelength $1100 \,\mathring{A}$ are used?

Photoelectric effect can be explained by

When a photon of energy $6 \, eV$ is incident on a metal surface,the maximum kinetic energy of the emitted photoelectrons is $4 \, eV$. The value of the stopping potential is .......... $V$.