PHYSICS
PHYSICS
5th Edition
ISBN: 2818440038631
Author: GIAMBATTISTA
Publisher: MCG
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Question
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Chapter 27, Problem 61P

(a)

To determine

The Planck’s constant from the graph given in question.

(a)

Expert Solution
Check Mark

Answer to Problem 61P

The Planck’s constant from the graph given in question is 6.66×1034Js.

Explanation of Solution

Write Einstein’s photoelectric effect.

  Kmax=hfϕ                                                                                                         (I)

Here, Kmax is the maximum kinetic energy of the electron, h is the Planck’s constant, f is the frequency of the light and ϕ is the work function of the metal.

Write the expression for the kinetic energy of the electron.

  Kmax=eVs

Here, e is the electronic charge, Vs is the stopping potential .

Substitute eVs for Kmax in equation (I) to get linear equation between stopping potential and frequency of light.

  eVs=hfϕVs=hefϕe                                                                                                            (II)

Write the general equation of straight line in xy graph.

  y=mx+c                                                                                                             (III)

Here, m is the slope of xy graph and c is the y intercept.

Equation (II) and (III) indicates that, Vs versus f is a straight line with slope he and ϕe as y intercept of the graph.

Write the expression for the slope of straight line in stopping potential versus frequency graph.

  m=V2V1f2f1                                                                                                          (IV)

Here, V1,V2 are the stopping potential at f1 and f2 respectively.

Conclusion:

The slope of stopping potential versus frequency graph given in question gives that value of h.

From graph, stopping potential at 800THz is 1.5V and stopping potential at 439THz is 0V.

Substitute 1.5V for V2, 0V for V1 , 800THz for f2 and 439THz for f1 in equation (IV) to get m.

  m=1.5V0V(800THz439THz)×1012Hz1THz=4.155×1015Vs

It is found that slope is equal to he.

Equate slope obtained with he to get h.

  he=4.155×1015Vsh=e(4.155×1015Vs)

Substitute 1.602×1019C in above equation to get h.

  h=(1.602×1019C)(4.155×1015Vs)=6.66×1034Js

Therefore, the Planck’s constant from the graph given in question is 6.66×1034Js.

(b)

To determine

The work function of the metal from the data.

(b)

Expert Solution
Check Mark

Answer to Problem 61P

The work function of the metal from the data is 1.82eV .

Explanation of Solution

Rewrite equation (II) .

  Vs=hefϕe

In graph, the straight-line intercept the x axis at its threshold frequency. At this frequency stopping potential is zero indicating that kinetic energy of the electron is zero.

The threshold frequency of the metal from the graph is 439THz.

Substitute 0 for Vs , f0 for f and m for he in above equation to get equation of work function.

  0=mf0ϕeϕ=emf0                                                                                                           (V)

Here,f0 is the work function of the metal.

Conclusion:

Substitute 4.155×1015Vs for m , 439THz for f0 and 1.602×1019C fro e in equation (V) to get ϕ.

  ϕ=(1.602×1019C)(4.155×1015Vs)(439THz×1012Hz1THz)×1eV1.602×1019J=1.82eV

Therefore, the work function of the metal from the data is 1.82eV .

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Chapter 27 Solutions

PHYSICS

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