Question

If the work function of a metal is 6.63 eV, then find the threshold frequency for photoelectric effect.

• A. 1.9 × 10¹⁵ Hz
• B. 1.6×10¹⁵ Hz
• C. 2 x 10¹⁶ Hz
• D. 1.2×10¹⁵ Hz

Answer 1

The Correct Option is B

The threshold frequency (\(f_0\)) for the photoelectric effect is determined using the work function (\(\phi\)) and Planck's constant (\(h\)) via the relation \(\phi = h \cdot f_0\). In this equation:

• \(\phi\) represents the work function of the metal in electron volts (eV).
• \(h\) is Planck's constant, approximately \(4.1357 \times 10^{-15} \, \text{eV} \cdot \text{s}\).
• \(f_0\) is the threshold frequency to be calculated.

Given values:

• \(\phi = 6.63 \, \text{eV}\)
• \(h = 4.1357 \times 10^{-15} \, \text{eV} \cdot \text{s}\)

Substituting these values into the equation yields:

\(6.63 = 4.1357 \times 10^{-15} \cdot f_0\)

Solving for \(f_0\):

\(f_0 = \frac{6.63}{4.1357 \times 10^{-15}} \, \text{Hz}\)

The computed result is:

\(f_0 = 1.602 \times 10^{15} \, \text{Hz}\)

Rounded, this value is \(1.6 \times 10^{15} \, \text{Hz}\).

Consequently, the threshold frequency for the photoelectric effect is 1.6 × 10¹⁵ Hz.