Advertisements
Advertisements
प्रश्न
Use Einstein's photoelectric equation to explain the observations from this graph ?
Advertisements
उत्तर
The Einstein's photoelectric equation is \[K_{max} = \frac{1}{2}m \nu^2_{max} = h\nu - \phi_o\]
From the graph, we noticed that
(i) The value of stopping potential is different for radiation of different frequencies.
\[\text { As } \]
\[e V_o = h\nu - \phi_o \]
\[\text { Hence }, V_o \propto \nu\]
(ii) The value of stopping potential is more negative for radiation of higher incident frequency.
APPEARS IN
संबंधित प्रश्न
The photoelectric work function for a metal surface is 2.3 eV. If the light of wavelength 6800A is incident on the surface of metal, find threshold frequency and incident frequency. Will there be an emission of photoelectrons or not?
[Velocity of light c = 3 x 108 m/s,
Planck’s constant, h = 6.63 * 10-34 Js ]
Sketch the graphs showing variation of stopping potential with frequency of incident radiations for two photosensitive materials A and B having threshold frequencies vA > vB.
(i) In which case is the stopping potential more and why?
(ii) Does the slope of the graph depend on the nature of the material used? Explain.
The photoelectric work function for a metal is 4.2 eV. If the stopping potential is 3V, find the threshold wavelength and maximum kinetic energy of emitted electrons.
(Velocity of light in air = 3 x 108m/s,
Planck's constant = 6·63 x10-34 J -s,
Charg.e ori electron = 1·6 x 10 -19 C)
Light of intensity ‘I’ and frequency ‘v’ is incident on a photosensitive surface and causes photoelectric emission. What will be the effect on anode current when (i) the intensity of light is gradually increased. In each case, all other factors remain the same. Explain, giving justification in each case.
Light of intensity ‘I’ and frequency ‘v’ is incident on a photosensitive surface and causes photoelectric emission. What will be the effect on anode current when (ii) the frequency of incident radiation is increased. In each case, all other factors remain the same. Explain, giving justification in each case.
Light of intensity ‘I’ and frequency ‘v’ is incident on a photosensitive surface and causes photoelectric emission. What will be the effect on anode current when the anode potential is increased? In each case, all other factors remain the same. Explain, giving justification in each case.
The work functions for potassium and caesium are 2.25 eV and 2.14 eV respectively. Is the photoelectric effect possible for either of them if the incident wavelength is 5180 Å?
[Given : Planck’s constant = 6.63 x 10–34 J.s.;
Velocity of light = 3 x 108 m/s; 1 eV = 1.6 x 10–19 J]
What is photoelectri effect ? Defin (i) Stopping potential (ii) Photoelectric work function.
Plot a labelled graph of IVsl where Vs is stopping potential versus frequency f of the incident radiation.
State how will you use this graph to detennine the value of Planck's constant.
