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kykrilka [37]
4 years ago
9

When light of wavelength 240 nm falls on a potassium surface, electrons having a maximum kinetic energy of 2.93 eV are emitted.

Find values for the following.When light of wavelength 240 nm falls on a potassium surface, electrons having a maximum kinetic energy of 2.93 eV are emitted. Find values for the following. (a) the work function of potassium
eV

(b) the cutoff wavelength
nm

(c) the frequency corresponding to the cutoff wavelength
Hz
Physics
1 answer:
olchik [2.2K]4 years ago
7 0

A) We want to find the work function of the potassium. Apply this equation:

E = 1243/λ - Φ

E = energy of photoelectron, λ = incoming light wavelength, Φ = potassium work function

Given values:

E = 2.93eV, λ = 240nm

Plug in and solve for Φ:

2.93 = 1243/240 - Φ

Φ = 2.25eV

B) We want to find the threshold wavelength, i.e. find the wavelength such that the energy E of the photoelectrons is 0eV. Plug in E = 0eV and Φ = 2.25eV and solve for the threshold wavelength λ:

E = 1243/λ - Φ

0 = 1243/λ - Φ

0 = 1243/λ - 2.25

λ = 552nm

C) We want to find the frequency associated with the threshold wavelength. Apply this equation:

c = fλ

c = speed of light in a vacuum, f = frequency, λ = wavelength

Given values:

c = 3×10⁸m/s, λ = 5.52×10⁻⁷m

Plug in and solve for f:

3×10⁸ = f(5.52×10⁻⁷)

f = 5.43×10¹⁴Hz

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Three people pull simultaneously on a stubborn donkey. Jack pulls directly ahead of the donkey with a force of 63.9 N63.9 N , Ji
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Answer:

a) F = (137.4 i ^ + 185 j ^) N

b)    F = 230.2 N  ,  θ = 53.5º

Explanation:

In this exercise we ask to find the net force, for which we will define a coordinate system fix the donkey and use trigonometry to decompose the forces

Jack       F₁ₓ = 63.9 N

Jill          F₂ = 79.1 N with direction 45º to the left

              cos (180 -45) = F₂ₓ / F₂

              sin 135 = F_{2y} / F₂

              F₂ₓ = F₂ cos 135

              F_{2y} = F₂ sin 135

              F₂ₓ = 79.1 cos 135 = -55.9 N

              F_{2y} = 79.1 sin 135 = 55.9 N

Jane      F₃ = 183 N direction 45th to the right

             cos 45 = F₃ₓ / F3

             sin 45 = F_{3y} / F3

             F₃ₓ = F₃ cos 45 = 183 cos 45

             F_{₃y} = F₃ sin 45 = 183 sin 45

             F₃ₓ = 129.4 N

             F_{3y} = 129.4 N

we add each component of the force

       Fₓ = F₁ₓ + F₂ₓ + F₃ₓ

       Fₓ = 63.9 + (-55.9) + 129.4

       Fₓ = 137.4 N

       F_{y} = F_{2y} + F_{3y}

       F_{2y} = 55.9 + 129.4

       F_{2y} = 185.3 N

we can give the result of the forms

 

a) F = (137.4 i ^ + 185 j ^) N

b) in the form of module and angle

         F = RA (Fₓ² + F_{y}²)

         F = Ra (137² + 185²)

         F = 230.2 N

         tan θ = F_{y} / Fₓ

         θ = tan⁻¹ F_{y} / Fₓ

        θ = tan⁻¹ (185/137)

        θ = 53.5º

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