When a metal plate is illuminated with light of wavelengths $400 \ nm$ and $250 \ nm$,the maximum velocities of the emitted photoelectrons are $v$ and $2v$,respectively. The work function of the metal is ($h$ = Planck's constant; $c$ = speed of light in vacuum):

$A$ stream of photons having energy $3 \,eV$ each impinges on a potassium surface. The work function of potassium is $2.3 \,eV$. The emerging photo-electrons are slowed down by a copper plate placed $5 \,mm$ away. If the potential difference between the two metal plates is $1 \,V$,the maximum distance the electrons can move away from the potassium surface before being turned back is .......... $mm$.

The maximum velocity of the photoelectron emitted by the metal surface is $v$. The charge and mass of the photoelectron are denoted by $e$ and $m$ respectively. The stopping potential in volts is:

Let $v_1$ and $v_2$ be the maximum velocities of the emitted electrons when the surface of a metal is illuminated with light waves of energy $E_1 = 4 \text{ eV}$ and $E_2 = 2.5 \text{ eV}$,respectively. If the work function of the metal is $2 \text{ eV}$,then the ratio $\frac{v_1}{v_2}$ is

If the work function of a photometal is $6.825 \; eV$,its threshold wavelength will be ............. $\mathring A$ $(c = 3 \times 10^8 \; m/s)$.

Photons of energy $7 \,eV$ are incident on two metals $A$ and $B$ with work functions $6 \,eV$ and $3 \,eV$,respectively. The minimum de-Broglie wavelengths of the emitted photoelectrons with maximum kinetic energies are $\lambda_A$ and $\lambda_B$ respectively,where $\lambda_A / \lambda_B$ is nearly