Physics
Physics
5th Edition
ISBN: 9781260486919
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 27, Problem 94P

(a)

To determine

The work function of the metal.

(a)

Expert Solution
Check Mark

Answer to Problem 94P

The work function of the metal is 4.7eV_.

Explanation of Solution

Given that the wavelength of light is 180nm, the magnetic field strength is 7.5×105T, and the radius of the circular path is 6.7cm

Write the expression for the magnitude of the magnetic force on the electron when it enters perpendicular into the field.

  F=evB                                                                                                                  (I)

Here, F is the magnetic force, e is the charge of electron, v is the speed of electron, and B is the magnetic field strength.

The magnetic force is providing the necessary centripetal force for the motion of the electron. Thus, the magnetic force can be equated to the centripetal force on electron.

  evB=mv2r                                                                                                           (II)

Here, m is the mass of electron, and r is the radius of the circular path.

Solve equation (II) for v.

  v=eBrm                                                                                                               (III)

Write the expression for the maximum kinetic energy of the ejected electron.

  Kmax=12mv2                                                                                                      (IV)

Here, Kmax is the maximum kinetic energy of the ejected electron.

Write the expression for the maximum kinetic energy of the ejected electron in terms of the work function of the metal.

  Kmax=hcλϕ                                                                                                    (V)

Here, c is the speed of light,λ is the  h is the Planck’s constant, and ϕ is the work function of the metal.

Equate the right hand sides of equations (IV) and (V) and solve for ϕ.

  ϕ=hcλ12mv2                                                                                                     (VI)

Use equation (III) in (VI).

  ϕ=hcλ12m(eBrm)2=hcλ(eBr)22m                                                                                           (VI)

Conclusion:

Substitute 180nm for λ, 6.626×1034Js for h, 3.00×108m/s for c, 1.602×1019C for e, 7.5×105T for B, 6.7cm for r, and 9.1×1031kg for m in equation (VI) to find ϕ.

  ϕ=(6.626×1034Js)(3.00×108m/s)180nm[(1.602×1019C)(7.5×105T)(6.7cm)]22(9.1×1031kg)=(6.626×1034Js)(3.00×108m/s)180nm×1m1×109nm[(1.602×1019C)(7.5×105T)(6.7cm×1m100cm)]22(9.1×1031kg)=7.4833×1019J×1eV1.602×1019J=4.7eV

Therefore, the work function of the metal is 4.7eV_.

(b)

To determine

Whether the electrons with maximum kinetic energy follow a path with the maximum or minimum radius.

(b)

Expert Solution
Check Mark

Answer to Problem 94P

The electrons with maximum kinetic energy follow a path with the maximum radius.

Explanation of Solution

Equation (IV) indicates that the maximum kinetic energy of the electron is directly proportional to the square of the speed of the electrons.

  Kv2

Equation (III) indicates that the speed of the electron is directly proportional to the radius of the circular path.

  vr

Hence, it can be summarized that the maximum kinetic energy of the electron corresponds to the large value of the radius. Thus, The electrons with maximum kinetic energy follow a path with the maximum radius.

Conclusion:

Therefore, the electrons with maximum kinetic energy follow a path with the maximum radius.

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