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

To find the threshold frequency for the photoelectric effect, we can use the relationship between the work function (\(\phi\)) and the threshold frequency (\(f_0\)). The formula that relates these quantities is based on Einstein's photoelectric equation:

\(\phi = h \cdot f_0\) 

where:

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

Given:

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

Plugging in these values into the equation:

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

Solving for \(f_0\):

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

Calculating the division gives:

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

Rounding off, this corresponds to \(1.6 \times 10^{15} \, \text{Hz}\).

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

Answer 2

The threshold frequency \( \nu_0 \) is given by:

\(\phi_0 = h \nu_0\)

**Given:**
- \( \phi_0 = 6.63 \, \text{eV} = 6.63 \times 1.6 \times 10^{-19} \, \text{J} = 1.06 \times 10^{-18} \, \text{J} \),
- Planck’s constant \( h = 6.63 \times 10^{-34} \, \text{J·s} \).

Thus, the threshold frequency is:

\(\nu_0 = \frac{\phi_0}{h} = \frac{1.06 \times 10^{-18}}{6.63 \times 10^{-34}} \approx 1.6 \times 10^{15} \, \text{Hz}.\)

The correct option is (B) : 1.6×10¹⁵ Hz