What is the maximum kinetic energy of photoelectrons emitted when light of frequency 8 x 10^14 Hz is incident on a metal with work function 3 eV?
Practice Questions
1 question
Q1
What is the maximum kinetic energy of photoelectrons emitted when light of frequency 8 x 10^14 Hz is incident on a metal with work function 3 eV?
1 eV
3 eV
5 eV
7 eV
Maximum kinetic energy (K.E.) = hν - Φ = (4.14 x 10^-15 eV·s)(8 x 10^14 Hz) - 3 eV = 5 eV.
Questions & Step-by-step Solutions
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Q
Q: What is the maximum kinetic energy of photoelectrons emitted when light of frequency 8 x 10^14 Hz is incident on a metal with work function 3 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.
Steps: 8
Step 1: Identify the frequency of the light, which is given as 8 x 10^14 Hz.
Step 2: Identify the work function (Φ) of the metal, 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 incoming photons (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 units).
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.
