When radiation of wavelength $\lambda$ is incident on a photocell,the maximum velocity of the photoelectrons is $v$. What will be the maximum velocity of the photoelectrons when radiation of wavelength $3\lambda/4$ is incident on the photocell?

The work functions of metals $A, B$,and $C$ are $1.92 \ eV, 2.0 \ eV$,and $5 \ eV$ respectively. According to Einstein's photoelectric equation,which metal$(s)$ will emit photoelectrons when illuminated by light of wavelength $4100 \ \mathring{A}$?

Photons of wavelength $\lambda$ emitted by a source of power $P$ are incident on a photocell. If the current produced in the cell is $I$,then the percentage of incident photons which produce current in the photocell is: (where $h$ is Planck's constant and $c$ is the speed of light in vacuum)

When a metal surface is illuminated by light of wavelength $\lambda_1$ and $\lambda_2$,the maximum velocities of photoelectrons ejected are $V$ and $2V$ respectively. The work function of the metal is ($h=$ Planck's constant,$c=$ velocity of light,$\lambda_1 > \lambda_2$).

$A$ photoelectric surface is illuminated successively by monochromatic light of wavelength $\lambda$ and $\lambda /2$. If the maximum kinetic energy of the emitted photoelectrons in the second case is $3$ times that in the first case,the work function of the surface of the material is
$(h =$ Planck's constant,$c =$ speed of light $)$

An electromagnetic wave of wavelength $\lambda$ is incident on a photosensitive surface of negligible work function. If the photoelectron emitted from the surface has mass $m$ and de-Broglie wavelength $\lambda_{d}$,then