Einstein's photoelectric equation states that ${E_k} = h\nu - \phi$. In this equation,${E_k}$ refers to:

In a photoemissive cell with exciting wavelength $\lambda$,the fastest electron has speed $v$. If the exciting wavelength is changed to $\frac{3\lambda}{4}$,the speed of the fastest emitted electron will be

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 $eV$)

When light of frequencies ${\nu _1}$ and ${\nu _2}$ $({\nu _1} > {\nu _2})$ is incident on a metal surface,the ratio of the maximum kinetic energy of the emitted photoelectrons is $1:k$. What is the threshold frequency of the metal?

In a photoelectric experiment, the wavelength of the light incident on the metal is changed from $200 \, nm$ to $400 \, nm$. The decrease in the stopping potential is close to [Use $hc = 1240 \, eV \cdot nm$ where $h$ is Planck's constant and $c$ is the velocity of light]. (in $ \, V$)

If the stopping potential of photoelectrons emitted by a source placed at a distance of $0.2 \ m$ from a photocell is $0.6 \ V$,then the stopping potential of photoelectrons emitted by the same source placed at a distance of $0.6 \ m$ from the photocell will be .......... $V$.