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kykrilka [37]
3 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]3 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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The time when the particle is at rest is at 1.63 s or 3.36 s.

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<h3>When is a particle at rest?</h3>
  • A particle is at rest when the initial velocity of the particle is zero.

The time when the particle is at rest is calculated as follows;

s(t) = 2t³ - 15t² + 33t + 17

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The velocity is positive at when the time of motion is as follows;

0.

The total distance traveled in the first 10 seconds is calculated as follows;

2(10)^3 - 15(10)^2 + 33(10) + 17 = 847 \ m

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