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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

Respuesta :

Answer:

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

b) v = 6.31 * 10^5   m/s

c) Lambda = 195 nm

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

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