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Ghella [55]
3 years ago
6

The light intensity incident on a metallic surface with a work function of 1.88 eV produces photoelectrons with a maximum kineti

c energy of 1.2 eV.
if the frequency of the light is doubled, what is the maximum kinetic energy of the photoelectrons?
Answer in units of eV.

​
Physics
1 answer:
ivanzaharov [21]3 years ago
7 0

Answer:

KE = KE (incidental) - KE of emitted photons

or KE = h * f - Wf

So   h * f = KE + Wf = 1.2 + 1.88 = 3.08    incident energy

If you double the frequency then h * f = 6.16

KE = 6.16 - 1.2 = 4.96 eV

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The electric potential in a region of space is \[V=350/\sqrt{x ^{2}+y ^{2}}\] where x and y are in meters. what is the strength
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3 years ago
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Photons of wavelength 65.0 pm are Compton-scattered from a free electron which picks up a kinetic energy of 0.84 keV from the co
ElenaW [278]

Answer:

λ  = 65.6 pm

Explanation:

Given that

λo = 65 pm

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E_o=\dfrac{hC}{\lambda_0 }

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E=\dfrac{hC}{\lambda}

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3 0
3 years ago
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