The work function for a metal is `40 eV`. To emit photo electrons of zero velocity from the surface of the metal the wavelength of incident light should be `x nm`.
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The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength 400 mm is 500 mV. When the incident wavelength is
Correct Answer - A Energy of photon (1) `E_(1)=(12400)/(lambda(Å)) eV` I case `E_(1)=(12400)/(4000)=3.1eV` `:. V_("stopping")=500mrArrKE ` of emitted `bare=0.5`eV `:. E_("photon")=W+KEbare` `rArr` Work function w=2.6 eV II case `V_("stopping")=800mvrArrKE_(e^(-))=0.5eV` `:....
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A photo cell is illuminated by a small bright source placed 1m away When the same source of light is placed 2m away, the electrons emitted by photo ca
Correct Answer - d `Iprop(1)/(d^(2))` When source is placed 2m away then `I=I//4` The number of electrons emitted is derectly proportional to intensity .Hence number of emitted electrons is reduced to...
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When light of wavelength `lambda` is incident on a metal surface, stopping potential is found to be x. When light of wavelength n`lambda` is incident
Let `lambda_(0)` is the threshold wavelength. The work function is `phi = (hc)/(lambda_(0))` Now, by photoelectric equation `ex = (hc)/(lambda) - (hc)/(lambda_(0))` ……(i) `(ex)/(n+1) = (hc)/(nlambda) - (hc)/(lambda_(0))`……..(ii) From (i)...
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A meacury arec lamp provides `0.1` watt of ultra-violet radiation at a wavelength of `lambda=2537 Å` only. The photo tube (cathode of photo electric d
Correct Answer - A UV energy flux at a distance of `1m = (0.1)/(4pi xx 1^(2))` cross section (effective area) of atom `= pi xx (2 xx 10^-10)^(2) = 4pi xx...
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The stopping potential for the photo-electrons emitted from a metal surface of work-function 1.7 eV is 10.4 eV. Find the wavelength of the radiation u
Correct Answer - A::C `K_(max) = 10.4 eV , phi = 1.7 eV` `E = 10.4 + 1.7 = 12.1 eV rArr n = 3 ` to `n = 1` `12.1...
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Light of wavelength `lambda` strikes a metal surface with intensity `X` and the metal emits `Y` electrons per second of average energy `Z`. What will
Correct Answer - 1 Number of emitted electron `prop` Intensity of incident light.
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The work function of Cs is 2.14eV.Find (a) threshold frequency for Cs (b) Wavelength of incident light if the photo current is brought to zero by stop
Wave function `omega = 2.14eV` (a) Threshold frequency `omega = hnu_(0)` `nu_(0) = (omega)/(h) = (2.14 xx 1.6 xx 10^(-19))/(6.62 xx 10^(-34))` `=5.174 xx 10^(14)H_(2)` (b) `"As "k_(max) = eV_(0)=...
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Photoelectric work function for a surface is 2.4 eV. Light of wavelength `6.8 xx 10^(-7)`m shines on the surface. Find the frequency of incident light
Correct Answer - `v =4.41 xx 10^(14)Hz,v_(0)=5.82 xx 10^(14) Hz`, no photoelectric emission occurs as `v lt v_(0)`
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Calculate the kinetic energy of a photoelectron (in eV) emitted on shining light of wavelength `6.2 xx 10^(-6)m` on a metal surface. The work function
Correct Answer - 0.1 eV
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When light of wavelength 400nm is incident on the cathode of photocell, the stopping potential recorded is 6V. If the wavelength of the incident light
Correct Answer - 4.97 V
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