(i) In the explanation of photoeletric effect, we assume one photon of frequency v collides with an electron and transfers its energy. This leads . to the equation for the maximum energy Emaxof the emitted electron as Emax = hv – ɸ0
where ɸ0 is the work function of the metal. If an electron absorbs 2 photons (each of frequency v), what will be the maximum energy for the emitted electron?
(ii) Why is this fact (two photon absorption) not taken into consideration in our discussion of the stopping potential?
(i) In the explanation of photoeletric effect, we assume one photon of frequency v collides with an electron and transfers its energy. This leads . to the equation for the maximum energy Emaxof the emitted electron as Emax = hv – ɸ0
where ɸ0 is the work function of the metal. If an electron absorbs 2 photons (each of frequency v), what will be the maximum energy for the emitted electron?
(ii) Why is this fact (two photon absorption) not taken into consideration in our discussion of the stopping potential?
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1 Answer
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This is a Short Answer Type Questions as classified in NCERT Exemplar
Explanation- (i) Here it is given that, an electron absorbs 2 photons each of frequency ν then ν where, v′ is the frequency of emitted electron.
Given, Emax= hv-
Now, maximum energy for emitted electrons is Emax= h2v-
(ii) The probability of absorbing 2 photons by the same electron is very low. Hence, such emission will be negligible
Similar Questions for you
Based on theory
z² × (13.6) (1 - ¼) = 3 × (13.6)
z = 2 . (i)
h/√2mk? = (1/2.3) × h/√2mk?
=> k? = (2.3)²k? = 5.25k? (ii)
Now, k? = E? - Φ
k? = E? - Φ = z²E? - Φ
∴ k? /k? = (10.2 - Φ)/ (4 × 10.2 - Φ) = 1/5.25
=> Φ = 3eV
- (i)
- (ii)
from (i) & (ii)
ev
hu = hu0 + K.E
Cases u = 2u0
h2u0 = hu0 + K.E1
K.E1 = hu0
- (1)
Now, cases 2
h 5u0 = hu0 + k.E2
k.E2 = 4hu0
v2 =
v2 = 2v1
This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:
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