$A$ metal surface is illuminated by light of a given intensity and frequency, causing photoelectric emission. If the intensity of the incident light is reduced to $1/4$ of its original value, what will be the kinetic energy of the emitted photoelectrons?

The work function of a surface of a photosensitive material is $6.2\, eV$. The wavelength of the incident radiation for which the stopping potential is $5\, V$ lies in the

An electron moving with velocity $1.6 \times 10^7 \ m/s$ has a wavelength of $0.4 \ \mathring{A}$. The required accelerating voltage for the electron motion is (charge on electron $= 1.6 \times 10^{-19} \ C$,mass of electron $= 9 \times 10^{-31} \ kg$).

Statement $1$: $A$ metal surface is irradiated by monochromatic light of frequency $v > v_0$. The maximum kinetic energy and stopping potential are $K_{max}$ and $V_0$ respectively. If the frequency of the incident light is doubled, then $K_{max}$ and $V_0$ will also be doubled. Statement $2$: The stopping potential and maximum kinetic energy of photoelectrons emitted from a surface depend linearly on the frequency of the incident light.

Explain: "For photoelectric emission, the frequency of incident radiation should be greater than the threshold frequency."

The work function of tungsten is $4.50 \, eV$. The wavelength of the fastest electron emitted when light whose photon energy is $5.50 \, eV$ falls on the tungsten surface is ......... $\mathring{A}$.