The work function of a metallic surface is $5.01\ eV$. Photoelectrons are emitted when light of wavelength $2000\ \mathring{A}$ falls on it. The minimum potential difference required to stop the fastest photoelectrons is ................. $V$.

When a metallic 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 metallic surface is:

$A$ certain metallic surface is illuminated with monochromatic light of wavelength $\lambda$. The stopping potential for the photoelectric current for this light is $3V_0$. If the same surface is illuminated with light of wavelength $2\lambda$,the stopping potential is $V_0$. The threshold wavelength for this surface for the photoelectric effect is:

The frequency of the incident light falling on a photosensitive metal plate is doubled. The kinetic energy of the emitted photoelectrons is:

Photoelectric emission occurs only when the incident light has more than a certain minimum

In an experiment on photoelectric emission from a metallic surface,the wavelength of incident light is $2 \times 10^{-7} \,m$ and the stopping potential is $2.5 \,V$. The threshold frequency of the metal (in $Hz$) is approximately (charge of electron $e=1.6 \times 10^{-19} \,C$,Planck's constant $h=6.6 \times 10^{-34} \,J-s$):