Question

A photon of wavelength \( 3000 \) Å strikes a metal surface. The work function of the metal is \( 2.13 \) eV. What is the kinetic energy of the emitted photoelectron? (\( h = 6.626 \times 10^{-34} \) Js)

• A. 4.0 eV
• B. 3.0 eV
• C. 2.0 eV
• D. 1.0 eV

Hint:

The kinetic energy of emitted electrons is the excess energy after overcoming the work function of the metal.

Answer 1

The Correct Option is C

Step 1: {Using Einstein's Photoelectric Equation}
\[ KE = h\nu - \phi \] Since \( \nu = \frac{c}{\lambda} \), we use: \[ E = \frac{hc}{\lambda} \] Step 2: {Substituting Values}
\[ E = \frac{(6.626 \times 10^{-34}) (3 \times 10^8)}{3 \times 10^{-7}} \] \[ E = 6.6 \times 10^{-19} J \] Converting to eV: \[ E = \frac{6.6 \times 10^{-19}}{1.6 \times 10^{-19}} = 4.125 eV \] Step 3: {Calculating Kinetic Energy}
\[ KE = 4.125 - 2.13 = 2.0 eV \] Thus, the correct answer is (C).

Answer 2

Step 1: Write the photoelectric equation
According to Einstein’s photoelectric equation:
Kinetic Energy (K.E.) = Energy of photon – Work function
K.E. = \( E - \phi \)

Step 2: Calculate the energy of the photon
The energy of a photon is given by:
\( E = \frac{hc}{\lambda} \)
Where:
- \( h = 6.626 \times 10^{-34} \) Js
- \( c = 3 \times 10^8 \) m/s
- \( \lambda = 3000 \) Å = \( 3000 \times 10^{-10} \) m = \( 3 \times 10^{-7} \) m

Now calculate:
\( E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{3 \times 10^{-7}} \)
\( E = \frac{1.9878 \times 10^{-25}}{3 \times 10^{-7}} = 6.626 \times 10^{-19} \) J
Convert to eV:
1 eV = \( 1.6 \times 10^{-19} \) J
So, \( E = \frac{6.626 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 4.14 \) eV

Step 3: Subtract the work function
Work function \( \phi = 2.13 \) eV
K.E. = \( 4.14 - 2.13 = 2.01 \) eV

Step 4: Final Answer
The kinetic energy of the emitted photoelectron is:
2.0 eV