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 eV
3 eV
5 eV
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.
