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What is the maximum kinetic energy of photoelectrons if the incident light has a

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Question: What is the maximum kinetic energy of photoelectrons if the incident light has a frequency of 8 x 10^14 Hz and the work function is 3 eV?

Options:

  1. 1 eV
  2. 3 eV
  3. 5 eV
  4. 7 eV

Correct Answer: 5 eV

Solution: 

Maximum kinetic energy (K.E.) = hν - Φ = (4.14 x 10^-15 eV·s)(8 x 10^14 Hz) - 3 eV = 5 eV.

What is the maximum kinetic energy of photoelectrons if the incident light has a

Practice Questions

Q1
What is the maximum kinetic energy of photoelectrons if the incident light has a frequency of 8 x 10^14 Hz and the work function is 3 eV?
  1. 1 eV
  2. 3 eV
  3. 5 eV
  4. 7 eV

Questions & Step-by-Step Solutions

What is the maximum kinetic energy of photoelectrons if the incident light has a frequency of 8 x 10^14 Hz and the work function is 3 eV?
  • Step 1: Identify the frequency of the incident light, which is given as 8 x 10^14 Hz.
  • Step 2: Identify the work function (Φ), which is given as 3 eV.
  • Step 3: Use the formula for maximum kinetic energy (K.E.) of photoelectrons: K.E. = hν - Φ.
  • Step 4: Find the value of Planck's constant (h), which is approximately 4.14 x 10^-15 eV·s.
  • Step 5: Calculate the energy of the incident light (hν) by multiplying Planck's constant (h) by the frequency (ν): hν = (4.14 x 10^-15 eV·s) * (8 x 10^14 Hz).
  • Step 6: Perform the multiplication: hν = 4.14 x 8 = 33.12 eV (after adjusting for the powers of 10).
  • Step 7: Now, substitute the values into the K.E. formula: K.E. = 33.12 eV - 3 eV.
  • Step 8: Calculate the maximum kinetic energy: K.E. = 30.12 eV.
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