When a photon of energy $hv$ is incident on an aluminum plate with a work function $E_0$,the maximum kinetic energy of the emitted photoelectrons is $K$. If the frequency of the incident radiation is doubled,the maximum kinetic energy of the emitted photoelectrons will be .......

$A$ metal surface is illuminated by light of a given intensity and frequency, causing photoelectric emission. If the intensity of the incident light is reduced to $1/4$ of its original value, what will be the kinetic energy of the emitted photoelectrons?

The maximum velocity of photoelectrons emitted from a metal surface is $1.2 \times 10^6 \ m/s$. If the specific charge of the electron is $1.8 \times 10^{11} \ C/kg$,then the value of the stopping potential in volts is:

$A$ silver sphere of radius $1 \ cm$ and work function $4.7 \ eV$ is suspended from an insulating thread in free space. It is under continuous illumination of $200 \ nm$ wavelength light. As photoelectrons are emitted,the sphere gets charged and acquires a potential. The maximum number of photoelectrons emitted from the sphere is $A \times 10^Z$ (where $1 < A < 10$). The value of $Z$ is:

$A$ photoelectric surface is illuminated successively by monochromatic light of wavelength $\lambda$ and $(\lambda / 3)$. If the maximum kinetic energy of the emitted photoelectrons in the second case is $4$ times that in the first case,the work function of the surface of the material is ($h=$ Planck's constant,$c=$ speed of light).

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.)