The work function of caesium is $2.14\, eV$. Find the wavelength of the incident light if the photo current is brought to zero by a stopping potential of $0.60\, V$. (Result in $nm$)

Light of wavelength $\lambda$ falls on a metal having work function $\frac{hc}{\lambda_0}$. Photoelectric effect will take place only if ($\lambda_0$ is the threshold wavelength).

If the surface of a metal is successively exposed to radiation of wavelengths $\lambda_1 = 350 \, nm$ and $\lambda_2 = 450 \, nm$,the maximum velocity of photoelectrons differs by a factor of $2$. The work function of this metal is:

$A$ light whose frequency is equal to $6 \times 10^{14} \, Hz$ is incident on a metal whose work function is $2 \, eV$. $[h = 6.63 \times 10^{-34} \, Js, 1 \, eV = 1.6 \times 10^{-19} \, J]$. The maximum kinetic energy of the emitted electrons will be ............ $eV$. (in $.49$)

Light of wavelength $4000 \,\mathring A$ falls on a photosensitive metal and a negative $2 \,V$ potential stops the emitted electrons. The work function of the material (in $eV$) is approximately $(h = 6.6 \times 10^{-34} \,Js, \,e = 1.6 \times 10^{-19} \,C, \,c = 3 \times 10^8 \,m/s)$.

The maximum velocity of the photoelectrons emitted by a metal surface is $9 \times 10^5 \ m/s$. The value of the ratio of charge $(e)$ to mass $(m)$ of the photoelectron is $1.8 \times 10^{11} \ C/kg$. The value of the stopping potential in volts is: