Physics, asked by g9racul0iabkokin, 1 year ago

By how much would the stopping potential for a given photosensitive surface go up if the frequency of the incident radiations were to be increased from 4 x 1015 Hz to 8 x 1015 Hz? Given h = 6.4 x 10-34 J-s, e = 1.6 x 10-19 C and c = 3 x 108 ms-1

Answers

Answered by kvnmurty
31
increase in energy of the photons = h (f2 - f1)
     = 6.4 * 10⁻³⁴ [8 * 10¹⁵ - 4 *10¹⁵]
     = 25.6 * 10⁻¹⁹ Joules

Increase in Stopping potential = 25.6 * 10⁻¹⁹ J /1.6 * 10⁻¹⁹ C
            = 16.0  Volts
 
Stopping potential is the voltage difference applied to the photoelectrons, to counter the KE given to them by the photons.  The electric field E = V/e , applied in opposite direction so that electrons stay inside the atoms and do not eject out of the surface.

The kinetic energy of electron is the energy in incident photons minus the escape energy of the electron in the atom.

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