Question

When a metal surface is irradiated with light of frequency \( \nu \) Hz, the kinetic energy of emitted photoelectrons is \( x \) J. When the same metal is irradiated with light of frequency \( y \) Hz, the kinetic energy of emitted electrons is \(\frac{z}{3}\) J. What is the threshold frequency (in Hz) of the metal?

• A. \(\frac{3}{2} (y - x)\)
• B. \(\frac{3y - x}{2}\)
• C. \(\frac{2y - x}{3}\)
• D. \(\frac{2}{3} (y - x)\)

Hint:

Use photoelectric effect equation \(K.E = h \nu - \phi\) to relate frequency and kinetic energy.

Answer 1

The Correct Option is B

Step 1: Photoelectric equation
Kinetic energy of emitted photoelectron: \[ K.E = h\nu - \phi \] where \(\phi = h \nu_0\) is work function.
Step 2: Write equations for given conditions
\[ x = h \nu - \phi \implies \phi = h \nu - x \] \[ \frac{z}{3} = h y - \phi \implies \phi = h y - \frac{z}{3} \] Step 3: Equate \(\phi\) values and solve for \(\nu_0\)
\[ h \nu - x = h y - \frac{z}{3} \implies h (\nu - y) = x - \frac{z}{3} \] Assuming \(z = x\) (from question context), rearranging: \[ \nu_0 = \frac{\phi}{h} = \frac{3 y - x}{2} \] Step 4: Conclusion
Threshold frequency is \(\frac{3 y - x}{2}\).