The longest wavelength of light that can initiate the photoelectric effect in a metal with a work function of $9 \ eV$ is:

$A$ light source is placed at a distance of $20 \ cm$ from a photocell,and the measured stopping potential is $0.6 \ V$. If the distance of the source is changed to $40 \ cm$,what will be the stopping potential in $V$?

The threshold wavelengths for photoelectric emission from two metals $A$ and $B$ are $400 \ nm$ and $800 \ nm$ respectively. The ratio of their work functions,$\phi_{A} : \phi_{B}$ is:

In a photocell,the incident wavelength is $\lambda$. The maximum speed of the emitted photoelectrons is $u$. If the incident wavelength is changed to $3\lambda / 4$,then the maximum speed of the emitted photoelectrons will be:

The variation of stopping potential for metals $A$,$B$,$C$ and $D$ with the frequency of incident radiation is shown in the figure. For which metal is the stopping potential higher for a given frequency of incident radiation $(v)$ if the threshold frequency is lower $(\nu_o)$?

Given that a photon of light of wavelength $10,000\;\mathring A$ has an energy equal to $1.23\; eV$. When light of wavelength $5000\;\mathring A$ and intensity $I_0$ falls on a photoelectric cell,the saturation current is $0.40 \times 10^{-6}\; A$ and the stopping potential is $1.36\; V$; then the work function is ..... $eV$.