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: {Apply Einstein's Photoelectric Equation}
\[KE = hu - \phi\]Since \( u = \frac{c}{\lambda} \), the equation becomes:\[E = \frac{hc}{\lambda}\]Step 2: {Input Values}
\[E = \frac{(6.626 \times 10^{-34}) (3 \times 10^8)}{3 \times 10^{-7}}\]\[E = 6.6 \times 10^{-19} J\]Convert to electron volts (eV):\[E = \frac{6.6 \times 10^{-19}}{1.6 \times 10^{-19}} = 4.125 eV\]Step 3: {Determine Kinetic Energy}
\[KE = 4.125 - 2.13 = 2.0 eV\]The solution is (C).