The value of the Planck’s constant and the work function of the metal.

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Answer 1

Question

Chapter 3, Problem 39P

To determine

The value of the Planck’s constant and the work function of the metal.

Expert Solution & Answer

Answer to Problem 39P

The value of the Planck’s constant is 6.93×10−34 J⋅s and the value of the work function of the metal is 6.37×10−19 J.

Explanation of Solution

Write the expression for the kinetic energy of the electron.

Kmax=eV0 (I)

Here, Kmax is the maximum kinetic energy of photon, e is the charge on an electron and V0 is the stopping potential.

Write the expression for the energy of the photon of particular wavelength.

E=hcλ (II)

Here, E is the energy of the photon, h is the Planck’s constant, c is the speed of light and λ is the wavelength of photon.

Write the expression for the work function of the metal.

E=ϕ+Kmax (III)

Here, ϕ is the work function of the metal.

Conclusion:

Substitute 1.6×10−19 C for e and 1.0 V for V0 in equation (I).

Kmax=(1.6×10−19 C)⋅(1.0 V)=1.6×10−19 J

Substitute 3×108 m/s for c and 260×10−9 m for λ in equation (II).

E=h(3×108 m/s)260×10−9 m=(1.15×1015)h J

Substitute (1.15×1015)h J for E and 1.6×10−19 J for Kmax in equation (III).

(1.15×1015)h J=ϕ+(1.6×10−19 J) (IV)

Substitute 1.6×10−19 C for e and 2.3 V for V0 in equation (I).

Kmax=(1.6×10−19 C)⋅(2.3 V)=3.68×10−19 J

Substitute 3×108 m/s for c and 207×10−9 m for λ in equation (II).

E=h(3×108 m/s)207×10−9 m=(1.45×1015)h J

Substitute (1.45×1015)h J for E and 3.68×10−19 J for Kmax in equation (III).

(1.45×1015)h J=ϕ+(3.68×10−19 J) (V)

Subtract the equation (IV) from (V).

0.3h×1015=2.08×10−19h=2.08×10−190.3×1015=6.93×10−34 J⋅s

Thus, the value of the Planck’s constant is 6.93×10−34 J⋅s.

Substitute 6.93×10−34 J⋅s for h in equation (IV).

(1.15×1015)(6.93×10−34 J⋅s) J=ϕ+(1.6×10−19 J)ϕ=(7.97×10−19)−(1.6×10−19)=6.37×10−19 J

Thus, the value of the Planck’s constant is 6.93×10−34 J⋅s and the value of the work function of the metal is 6.37×10−19 J.

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Answer 2

Question

Chapter 3, Problem 39P

To determine

The value of the Planck’s constant and the work function of the metal.

Expert Solution & Answer

Answer to Problem 39P

The value of the Planck’s constant is 6.93×10−34 J⋅s and the value of the work function of the metal is 6.37×10−19 J.

Explanation of Solution

Write the expression for the kinetic energy of the electron.

Kmax=eV0 (I)

Here, Kmax is the maximum kinetic energy of photon, e is the charge on an electron and V0 is the stopping potential.

Write the expression for the energy of the photon of particular wavelength.

E=hcλ (II)

Here, E is the energy of the photon, h is the Planck’s constant, c is the speed of light and λ is the wavelength of photon.

Write the expression for the work function of the metal.

E=ϕ+Kmax (III)

Here, ϕ is the work function of the metal.

Conclusion:

Substitute 1.6×10−19 C for e and 1.0 V for V0 in equation (I).

Kmax=(1.6×10−19 C)⋅(1.0 V)=1.6×10−19 J

Substitute 3×108 m/s for c and 260×10−9 m for λ in equation (II).

E=h(3×108 m/s)260×10−9 m=(1.15×1015)h J

Substitute (1.15×1015)h J for E and 1.6×10−19 J for Kmax in equation (III).

(1.15×1015)h J=ϕ+(1.6×10−19 J) (IV)

Substitute 1.6×10−19 C for e and 2.3 V for V0 in equation (I).

Kmax=(1.6×10−19 C)⋅(2.3 V)=3.68×10−19 J

Substitute 3×108 m/s for c and 207×10−9 m for λ in equation (II).

E=h(3×108 m/s)207×10−9 m=(1.45×1015)h J

Substitute (1.45×1015)h J for E and 3.68×10−19 J for Kmax in equation (III).

(1.45×1015)h J=ϕ+(3.68×10−19 J) (V)

Subtract the equation (IV) from (V).

0.3h×1015=2.08×10−19h=2.08×10−190.3×1015=6.93×10−34 J⋅s

Thus, the value of the Planck’s constant is 6.93×10−34 J⋅s.

Substitute 6.93×10−34 J⋅s for h in equation (IV).

(1.15×1015)(6.93×10−34 J⋅s) J=ϕ+(1.6×10−19 J)ϕ=(7.97×10−19)−(1.6×10−19)=6.37×10−19 J

Thus, the value of the Planck’s constant is 6.93×10−34 J⋅s and the value of the work function of the metal is 6.37×10−19 J.

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Answer 3

Question

Chapter 3, Problem 39P

To determine

The value of the Planck’s constant and the work function of the metal.

Expert Solution & Answer

Answer to Problem 39P

The value of the Planck’s constant is 6.93×10−34 J⋅s and the value of the work function of the metal is 6.37×10−19 J.

Explanation of Solution

Write the expression for the kinetic energy of the electron.

Kmax=eV0 (I)

Here, Kmax is the maximum kinetic energy of photon, e is the charge on an electron and V0 is the stopping potential.

Write the expression for the energy of the photon of particular wavelength.

E=hcλ (II)

Here, E is the energy of the photon, h is the Planck’s constant, c is the speed of light and λ is the wavelength of photon.

Write the expression for the work function of the metal.

E=ϕ+Kmax (III)

Here, ϕ is the work function of the metal.

Conclusion:

Substitute 1.6×10−19 C for e and 1.0 V for V0 in equation (I).

Kmax=(1.6×10−19 C)⋅(1.0 V)=1.6×10−19 J

Substitute 3×108 m/s for c and 260×10−9 m for λ in equation (II).

E=h(3×108 m/s)260×10−9 m=(1.15×1015)h J

Substitute (1.15×1015)h J for E and 1.6×10−19 J for Kmax in equation (III).

(1.15×1015)h J=ϕ+(1.6×10−19 J) (IV)

Substitute 1.6×10−19 C for e and 2.3 V for V0 in equation (I).

Kmax=(1.6×10−19 C)⋅(2.3 V)=3.68×10−19 J

Substitute 3×108 m/s for c and 207×10−9 m for λ in equation (II).

E=h(3×108 m/s)207×10−9 m=(1.45×1015)h J

Substitute (1.45×1015)h J for E and 3.68×10−19 J for Kmax in equation (III).

(1.45×1015)h J=ϕ+(3.68×10−19 J) (V)

Subtract the equation (IV) from (V).

0.3h×1015=2.08×10−19h=2.08×10−190.3×1015=6.93×10−34 J⋅s

Thus, the value of the Planck’s constant is 6.93×10−34 J⋅s.

Substitute 6.93×10−34 J⋅s for h in equation (IV).

(1.15×1015)(6.93×10−34 J⋅s) J=ϕ+(1.6×10−19 J)ϕ=(7.97×10−19)−(1.6×10−19)=6.37×10−19 J

Thus, the value of the Planck’s constant is 6.93×10−34 J⋅s and the value of the work function of the metal is 6.37×10−19 J.

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Answer 4

Question

Chapter 3, Problem 39P

To determine

The value of the Planck’s constant and the work function of the metal.

Expert Solution & Answer

Answer to Problem 39P

The value of the Planck’s constant is 6.93×10−34 J⋅s and the value of the work function of the metal is 6.37×10−19 J.

Explanation of Solution

Write the expression for the kinetic energy of the electron.

Kmax=eV0 (I)

Here, Kmax is the maximum kinetic energy of photon, e is the charge on an electron and V0 is the stopping potential.

Write the expression for the energy of the photon of particular wavelength.

E=hcλ (II)

Here, E is the energy of the photon, h is the Planck’s constant, c is the speed of light and λ is the wavelength of photon.

Write the expression for the work function of the metal.

E=ϕ+Kmax (III)

Here, ϕ is the work function of the metal.

Conclusion:

Substitute 1.6×10−19 C for e and 1.0 V for V0 in equation (I).

Kmax=(1.6×10−19 C)⋅(1.0 V)=1.6×10−19 J

Substitute 3×108 m/s for c and 260×10−9 m for λ in equation (II).

E=h(3×108 m/s)260×10−9 m=(1.15×1015)h J

Substitute (1.15×1015)h J for E and 1.6×10−19 J for Kmax in equation (III).

(1.15×1015)h J=ϕ+(1.6×10−19 J) (IV)

Substitute 1.6×10−19 C for e and 2.3 V for V0 in equation (I).

Kmax=(1.6×10−19 C)⋅(2.3 V)=3.68×10−19 J

Substitute 3×108 m/s for c and 207×10−9 m for λ in equation (II).

E=h(3×108 m/s)207×10−9 m=(1.45×1015)h J

Substitute (1.45×1015)h J for E and 3.68×10−19 J for Kmax in equation (III).

(1.45×1015)h J=ϕ+(3.68×10−19 J) (V)

Subtract the equation (IV) from (V).

0.3h×1015=2.08×10−19h=2.08×10−190.3×1015=6.93×10−34 J⋅s

Thus, the value of the Planck’s constant is 6.93×10−34 J⋅s.

Substitute 6.93×10−34 J⋅s for h in equation (IV).

(1.15×1015)(6.93×10−34 J⋅s) J=ϕ+(1.6×10−19 J)ϕ=(7.97×10−19)−(1.6×10−19)=6.37×10−19 J

Thus, the value of the Planck’s constant is 6.93×10−34 J⋅s and the value of the work function of the metal is 6.37×10−19 J.

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Answer 5

Chapter 3, Problem 39P

To determine

The value of the Planck’s constant and the work function of the metal.

Expert Solution & Answer

Answer to Problem 39P

The value of the Planck’s constant is 6.93×10−34 J⋅s and the value of the work function of the metal is 6.37×10−19 J.

Explanation of Solution

Write the expression for the kinetic energy of the electron.

Kmax=eV0 (I)

Here, Kmax is the maximum kinetic energy of photon, e is the charge on an electron and V0 is the stopping potential.

Write the expression for the energy of the photon of particular wavelength.

E=hcλ (II)

Here, E is the energy of the photon, h is the Planck’s constant, c is the speed of light and λ is the wavelength of photon.

Write the expression for the work function of the metal.

E=ϕ+Kmax (III)

Here, ϕ is the work function of the metal.

Conclusion:

Substitute 1.6×10−19 C for e and 1.0 V for V0 in equation (I).

Kmax=(1.6×10−19 C)⋅(1.0 V)=1.6×10−19 J

Substitute 3×108 m/s for c and 260×10−9 m for λ in equation (II).

E=h(3×108 m/s)260×10−9 m=(1.15×1015)h J

Substitute (1.15×1015)h J for E and 1.6×10−19 J for Kmax in equation (III).

(1.15×1015)h J=ϕ+(1.6×10−19 J) (IV)

Substitute 1.6×10−19 C for e and 2.3 V for V0 in equation (I).

Kmax=(1.6×10−19 C)⋅(2.3 V)=3.68×10−19 J

Substitute 3×108 m/s for c and 207×10−9 m for λ in equation (II).

E=h(3×108 m/s)207×10−9 m=(1.45×1015)h J

Substitute (1.45×1015)h J for E and 3.68×10−19 J for Kmax in equation (III).

(1.45×1015)h J=ϕ+(3.68×10−19 J) (V)

Subtract the equation (IV) from (V).

0.3h×1015=2.08×10−19h=2.08×10−190.3×1015=6.93×10−34 J⋅s

Thus, the value of the Planck’s constant is 6.93×10−34 J⋅s.

Substitute 6.93×10−34 J⋅s for h in equation (IV).

(1.15×1015)(6.93×10−34 J⋅s) J=ϕ+(1.6×10−19 J)ϕ=(7.97×10−19)−(1.6×10−19)=6.37×10−19 J

Thus, the value of the Planck’s constant is 6.93×10−34 J⋅s and the value of the work function of the metal is 6.37×10−19 J.

Answer 6

Chapter 3, Problem 39P

To determine

The value of the Planck’s constant and the work function of the metal.

Expert Solution & Answer

Answer to Problem 39P

The value of the Planck’s constant is 6.93×10−34 J⋅s and the value of the work function of the metal is 6.37×10−19 J.

Explanation of Solution

Write the expression for the kinetic energy of the electron.

Kmax=eV0 (I)

Here, Kmax is the maximum kinetic energy of photon, e is the charge on an electron and V0 is the stopping potential.

Write the expression for the energy of the photon of particular wavelength.

E=hcλ (II)

Here, E is the energy of the photon, h is the Planck’s constant, c is the speed of light and λ is the wavelength of photon.

Write the expression for the work function of the metal.

E=ϕ+Kmax (III)

Here, ϕ is the work function of the metal.

Conclusion:

Substitute 1.6×10−19 C for e and 1.0 V for V0 in equation (I).

Kmax=(1.6×10−19 C)⋅(1.0 V)=1.6×10−19 J

Substitute 3×108 m/s for c and 260×10−9 m for λ in equation (II).

E=h(3×108 m/s)260×10−9 m=(1.15×1015)h J

Substitute (1.15×1015)h J for E and 1.6×10−19 J for Kmax in equation (III).

(1.15×1015)h J=ϕ+(1.6×10−19 J) (IV)

Substitute 1.6×10−19 C for e and 2.3 V for V0 in equation (I).

Kmax=(1.6×10−19 C)⋅(2.3 V)=3.68×10−19 J

Substitute 3×108 m/s for c and 207×10−9 m for λ in equation (II).

E=h(3×108 m/s)207×10−9 m=(1.45×1015)h J

Substitute (1.45×1015)h J for E and 3.68×10−19 J for Kmax in equation (III).

(1.45×1015)h J=ϕ+(3.68×10−19 J) (V)

Subtract the equation (IV) from (V).

0.3h×1015=2.08×10−19h=2.08×10−190.3×1015=6.93×10−34 J⋅s

Thus, the value of the Planck’s constant is 6.93×10−34 J⋅s.

Substitute 6.93×10−34 J⋅s for h in equation (IV).

(1.15×1015)(6.93×10−34 J⋅s) J=ϕ+(1.6×10−19 J)ϕ=(7.97×10−19)−(1.6×10−19)=6.37×10−19 J

Thus, the value of the Planck’s constant is 6.93×10−34 J⋅s and the value of the work function of the metal is 6.37×10−19 J.

Answer 7

Chapter 3, Problem 39P

To determine

The value of the Planck’s constant and the work function of the metal.

Answer 8

Expert Solution & Answer

Answer to Problem 39P

The value of the Planck’s constant is 6.93×10−34 J⋅s and the value of the work function of the metal is 6.37×10−19 J.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Write the expression for the kinetic energy of the electron.

Kmax=eV0 (I)

Here, Kmax is the maximum kinetic energy of photon, e is the charge on an electron and V0 is the stopping potential.

Write the expression for the energy of the photon of particular wavelength.

E=hcλ (II)

Here, E is the energy of the photon, h is the Planck’s constant, c is the speed of light and λ is the wavelength of photon.

Write the expression for the work function of the metal.

E=ϕ+Kmax (III)

Here, ϕ is the work function of the metal.

Conclusion:

Substitute 1.6×10−19 C for e and 1.0 V for V0 in equation (I).

Kmax=(1.6×10−19 C)⋅(1.0 V)=1.6×10−19 J

Substitute 3×108 m/s for c and 260×10−9 m for λ in equation (II).

E=h(3×108 m/s)260×10−9 m=(1.15×1015)h J

Substitute (1.15×1015)h J for E and 1.6×10−19 J for Kmax in equation (III).

(1.15×1015)h J=ϕ+(1.6×10−19 J) (IV)

Substitute 1.6×10−19 C for e and 2.3 V for V0 in equation (I).

Kmax=(1.6×10−19 C)⋅(2.3 V)=3.68×10−19 J

Substitute 3×108 m/s for c and 207×10−9 m for λ in equation (II).

E=h(3×108 m/s)207×10−9 m=(1.45×1015)h J

Substitute (1.45×1015)h J for E and 3.68×10−19 J for Kmax in equation (III).

(1.45×1015)h J=ϕ+(3.68×10−19 J) (V)

Subtract the equation (IV) from (V).

0.3h×1015=2.08×10−19h=2.08×10−190.3×1015=6.93×10−34 J⋅s

Thus, the value of the Planck’s constant is 6.93×10−34 J⋅s.

Substitute 6.93×10−34 J⋅s for h in equation (IV).

(1.15×1015)(6.93×10−34 J⋅s) J=ϕ+(1.6×10−19 J)ϕ=(7.97×10−19)−(1.6×10−19)=6.37×10−19 J

Thus, the value of the Planck’s constant is 6.93×10−34 J⋅s and the value of the work function of the metal is 6.37×10−19 J.

Answer 12

Ch. 3 - Prob. 1Q Ch. 3 - Prob. 2Q Ch. 3 - Prob. 3Q Ch. 3 - Prob. 4Q Ch. 3 - Prob. 5Q Ch. 3 - Prob. 6Q Ch. 3 - Prob. 7Q Ch. 3 - Prob. 8Q Ch. 3 - Prob. 9Q Ch. 3 - In the experiment of Example 3.2, how could you...

The value of the Planck’s constant and the work function of the metal.

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The value of the Planck’s constant and the work function of the metal.

Answer to Problem 39P

Explanation of Solution

Want to see more full solutions like this?

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