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When monochromatic light of an unknown wavelength falls on a sample of aluminum, a minimum potential of 2.27 V is required to stop all of the ejected photoelectrons. (The work function for aluminum is 4.08 eV.) HINT (a) Determine the maximum kinetic energy (in eV) of the ejected photoelectrons. eV (b) Determine the maximum speed (in m/s) of the ejected photoelectrons. m/s (c) Determine the wavelength in nm of the incident light. nm

User Bendik
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2 Answers

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Final answer:

To address the photoelectric effect in aluminum, we determined the maximum kinetic energy of photoelectrons (2.27 eV), discussed how to calculate their maximum speed, and outlined the method to find the wavelength of the incident monochromatic light using the stopping potential and the work function.

Step-by-step explanation:

The student is asking about the photoelectric effect in aluminum which involves concepts of physics. To solve these problems, one needs to use Einstein's photoelectric equation and basic principles of energy and momentum.

Maximum Kinetic Energy

The maximum kinetic energy (Kmax) of the photoelectrons can be determined using the equation Kmax = eV, where e is the charge of an electron and V is the stopping potential. Thus, Kmax = 2.27 eV.

Maximum Speed of Photoelectrons

To determine the maximum speed (vmax) of the photoelectrons, we can use the equation Kmax = (1/2)mv2, where m is the mass of an electron. By plugging in the values, we can solve for vmax.

Wavelength of Incident Light

The wavelength of the incident light can be found using Einstein's photoelectric equation, hf = φ + Kmax, where h is Planck's constant, f is the frequency of the light, and φ is the work function. With the given work function, and Kmax, we can determine the frequency and corresponding wavelength.

User Jon Lamb
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Answer:

a) KE max = 3.632 * 10^{-19}

b) v = 6.31 * 10^5 m/s

c) Lambda = 195 nm

Step-by-step explanation:

a) Work done is given by equation

W = V * q = change in kinetic energy = Final KE – Initial KE

Substituting the given values, we get –

V * 1.6*10^{-19} =0 - Initial KE

KE max = 2.27 * 1.6*10^{-19} = 3.632 * 10^{-19}

b) As we know KE = 0.5 mv^{2}

Substituting the given values, we get –

3.632 * 10^{-19} = 0.5 * (9.11 * 10^{-31}) v^2

v = 6.31 * 10^5 m/s

c) Incident energy = W + K max

Substituting the given values we get

hc/lambda = 4.08 * 1.6 * 10^{-19} J + 3.632 * 10^{-19} J

6.626 *10^{-34} * 3*10^8/lambda = 4.08 * 1.6 * 10^{-19} J + 3.632 * 10^{-19} J

Lambda = 1.95 * 10^7

Lambda = 195 nm

User CPak
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