Question

Radiation containing frequencies \(6 \times 10^{14}\,\text{Hz}\) and \(9 \times 10^{14}\,\text{Hz}\) falls on a metal surface with work function \(2.5\,\text{eV}\). Calculate the maximum energy of emitted electrons (in eV).

• A. \(0.8\)
• B. \(1.0\)
• C. \(1.2\)
• D. \(1.5\)

Hint:

In photoelectric effect problems, always use the highest incident frequency to calculate maximum kinetic energy.

Answer 1

The Correct Option is C

Step 1: Identify the relevant frequency.
The maximum kinetic energy of photoelectrons depends on the highest frequency of incident radiation.
Thus, we use: \[ \nu_{\max} = 9 \times 10^{14}\,\text{Hz}. \]
Step 2: Write the photoelectric equation.
\[ K_{\max} = h\nu - \phi, \] where \(h = 4.14 \times 10^{-15}\,\text{eV·s}\) and \(\phi = 2.5\,\text{eV}\).
Step 3: Calculate the photon energy.
\[ E = h\nu = (4.14 \times 10^{-15})(9 \times 10^{14}) = 3.73\,\text{eV}. \]
Step 4: Calculate maximum kinetic energy.
\[ K_{\max} = 3.73 - 2.5 = 1.23\,\text{eV}. \]
Step 5: Final answer (nearest suitable value).
\[ \boxed{K_{\max} \approx 1.2\,\text{eV}}. \]