The threshold wavelength for photoelectric emission from a metal surface is $5200 \ \mathring A$. Which of the following sources will cause photoelectric emission from this surface?

The surface of a metal is first illuminated with a light of wavelength $300 \ nm$ and later illuminated by another light of wavelength $500 \ nm$. It is observed that the ratio of maximum velocities of photoelectrons in the two cases is $3$. The work function of the metal is close to: (in $eV$)

The figure shows different graphs between stopping potential $(V_0)$ and frequency $(\nu)$ for photosensitive surfaces of cesium,potassium,sodium,and lithium. The plots are parallel. The correct ranking of the targets according to their work function,with the greatest first,is:

When a metal surface is illuminated by light of wavelength $\lambda_1$ and $\lambda_2$,the maximum velocities of photoelectrons ejected are $V$ and $2V$ respectively. The work function of the metal is ($h=$ Planck's constant,$c=$ velocity of light,$\lambda_1 > \lambda_2$).

The slope of the graph between stopping potential and the frequency of incident radiation for the photoelectric effect is . . . . . . .

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: