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kupik [55]
4 years ago
10

The light intensity incident on a metallic surface with a work function of 3 eV produces photoelectrons with a maximum kinetic e

nergy of 2 eV. The frequency of the light is doubled. Determine the maximum kinetic energy (in eV).
Physics
1 answer:
Vlad1618 [11]4 years ago
7 0

Answer:

7ev

Explanation:

The Energy of an incident light is given as E = hf;

Where h is Planks constant

f is frequency

Also.

E = Eo + K.E

Where Eo is work function and K.E is kinetic energy.

Hence substituting the given values,

E = 3ev + 2ev = 5ev

The frequency of the incident wave if it's doubled the Energy would be doubled since E is proportional to f with other elements been constant.

Hence :E = 10ev

But K.E = E -Eo

= 10ev - 3ev = 7ev

Maximum Kinetic Energy is 7ev.

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<h3><u>Given</u><u>:</u><u>-</u></h3>

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Harlamova29_29 [7]

Answer:

\lambda'=78.086\ nm

Explanation:

Given:

  • wavelength of light in the air, \lambda=530\times 10^{-9}\ m
  • time taken to travel from the source to the photocell via air, t=16.7\ s
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<u>Now we have the relation for time:</u>

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d=16.7\times 10^{-9}\times 3\times 10^8

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For the case when glass slab is inserted between the path of light:

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\frac{(5.01-0.87)}{3\times 10^8} +\frac{0.87}{v} =21.3\times 10^{-9}

v=4.42\times 10^7\ m.s^{-1}

Using Snell's law:

\frac{\lambda}{\lambda'} =\frac{c}{v}

\frac{530}{\lambda'} =\frac{3\times 10^8}{4.42\times 10^7}

\lambda'=78.086\ nm

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