When a certain metal surface is illuminated with light of wavelength $\lambda$,the stopping potential is $V$. When the same surface is illuminated by light of wavelength $2\lambda$,the stopping potential is $\frac{V}{3}$. The threshold wavelength for the surface is:

The work function of a material is $4.0 \ eV$. The longest wavelength of light that can cause the emission of photoelectrons from this material is approximately ............ $nm$.

The threshold wavelength for photoelectric emission from a material is $5500\,\mathring A$. Photoelectrons will be emitted when this material is illuminated with monochromatic radiation from a:
$A.$ $75\,W$ infra-red lamp
$B.$ $10\,W$ infra-red lamp
$C.$ $75\,W$ ultra-violet lamp
$D.$ $10\,W$ ultra-violet lamp
Choose the correct answer from the options given below:

$A$ photoelectric material having work-function $\phi_0$ is illuminated with light of wavelength $\lambda$ (where $\lambda < \frac{hc}{\phi_0}$). The fastest photoelectron has a de Broglie wavelength $\lambda_d$. $A$ change in wavelength of the incident light by $\Delta \lambda$ results in a change $\Delta \lambda_d$ in $\lambda_d$. Then the ratio $\frac{\Delta \lambda_d}{\Delta \lambda}$ is proportional to:

The potential difference that must be applied to stop the fastest photoelectrons emitted by a nickel surface,having work function $5.01 \, eV$,when ultraviolet light of $200 \, nm$ falls on it,must be ............... $V$.

Light of wavelength $488 \;nm$ is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the emitter, the stopping (cut-off) potential of photoelectrons is $0.38 \;V$. Find the work function (in $eV$) of the material from which the emitter is made.