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 threshold frequency for a photosensitive metal is $3.3 \times 10^{14} \text{ Hz}$. If light of frequency $8.20 \times 10^{14} \text{ Hz}$ is incident on this metal,the cut-off voltage for the photoelectron emission is nearly ............ $V$.

The threshold wavelength for photoelectric emission from a material is $5200 \, \mathring{A}$. Photo-electrons will be emitted when this material is illuminated with monochromatic radiation from a

Maximum velocity of the photoelectron emitted by a metal is $1.8 \times 10^{6} \ m/s$. Take the value of specific charge of the electron as $1.8 \times 10^{11} \ C/kg$. Then the stopping potential in volt is

If the energy of a photon is $25\, eV$ and the work function of the material is $7\, eV$,then the value of the stopping potential is :-................. $V$

When photons are incident on a photosensitive material of work function $1.5 \text{ eV}$,the maximum velocity of the emitted photoelectrons is $8 \times 10^5 \text{ m/s}$. The stopping potential of the photoelectrons is (Mass of the electron $= 9 \times 10^{-31} \text{ kg}$ and charge of the electron $= 1.6 \times 10^{-19} \text{ C}$) (in $V$)