Maximum velocity of photoelectrons emitted by a metal surface is $1.2 \times 10^6\, m/s$. Assuming the specific charge of the electron to be $1.8 \times 10^{11}\, C/kg$,the value of the stopping potential in volt will be

The de Broglie wavelength of the most energetic photoelectrons emitted from a photosensitive metal of work function $\phi$,when light of frequency $\nu$ is incident on it,is $\lambda$. Then $\nu =$ (where $h$ is Planck's constant and $m$ is the mass of the electron).

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

The correct graph between the maximum kinetic energy $(K_{\max})$ of a photoelectron and the inverse of the wavelength $(1/\lambda)$ of the incident radiation is given by the curve:

Yellow light of wavelength $557 \ nm$ is incident on a cesium surface. When the cathode-anode voltage drops below $0.25 \ V$,no photoelectrons flow in the circuit. What is the threshold wavelength of the cesium surface in $nm$?

The work function of a substance is $4.0 \,eV$. The longest wavelength of light that can cause photoelectron emission from this substance is approximately ......... $nm$.