If the work function of a metal is $\phi$ and the frequency of the incident light is $v$,there is no emission of photoelectrons for:

The electric field at the point associated with a light wave is given by $E = 200 [\sin(6 \times 10^{15} t) + \sin(9 \times 10^{15} t)] \, Vm^{-1}$. Given $h = 4.14 \times 10^{-15} \, eVs$. If this light falls on a metal surface having a work function of $2.50 \, eV$,the maximum kinetic energy of the photoelectrons will be ........... $eV$.

The graph of stopping potential $(V_0)$ versus the frequency $(f)$ of incident radiation for the photoelectric effect is ..........

If the threshold wavelength of light for photoelectric emission to take place from a metal surface is $6250 \ \text{Å}$, then the work function of the metal is (Planck's constant $= 6.6 \times 10^{-34} \ \text{Js}$) (in $\text{eV}$)

For a metal with a work function of $6.6 \text{eV}$, which of the following wavelengths of incident radiation does not cause the photoelectric effect (in $\text{nm}$)? (Take Planck's constant $h = 6.6 \times 10^{-34} \text{J s}$ and speed of light $c = 3 \times 10^8 \text{m/s}$)

$V$ (stopping potential) is plotted against $\frac{1}{\lambda}$,where $\lambda$ is the wavelength of incident radiation,for two metals.