Q. 83.6( 16 Votes )
The threshold frequency for a certain metal is 3.3 × 1014 Hz. If the light of frequency 8.2 × 1014 Hz is incident on the metal, predict the cutoff voltage for the photoelectric emission.
eV0 = hv - ϕ0 ……………… equation(1)
Where e = Charge on an electron = 1.6 × 10-19 C
V0 = Stopping Potential
h = plank’s constant
v = frequency of photon
ϕ0 = Work function of the metal
Threshold frequency is the minimum frequency at which photoemission starts taking place from a metal surface. It is represented by v0.
At threshold frequency, stopping potential is zero and energy of the emitted electron is also zero. Energy provided by the incident photon is just sufficient to eject out electrons from the metal surface.
ϕ0 = hv0 ....................equ.02
V0 = Threshold frequency
Putting value of ϕ0 from equ.02 to equ.01
eV0 = hv – hv0
∴ V0 = h (v-v0)/ e = 6.63 × 10-34 × (8.2-3.3) × 1014/1.6 × 10-19
= 2.03 V
For photo emission cut of potential is equal to 2.03 V
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If light of wavelength 412.5 nm is incident on each of the metals given below, which ones will show photoelectric emission and why ?
Physics - Board Papers
A. How does one explain the emission of electrons from a photosensitive surface with the help of Einstein’s photoelectric equation?
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