Question

The work function of a metal is 2 eV. What is the threshold frequency for the photoelectric emission? (Take Planck's constant $ h = 6.63 \times 10^{-34} \, \text{Js} $, $ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} $)

• A. \(4.8 \times 10^{14} \, \text{Hz}\)
• B. \(5.2 \times 10^{14} \, \text{Hz}\)
• C. \(6.2 \times 10^{14} \, \text{Hz}\)
• D. \(7.4 \times 10^{14} \, \text{Hz}\)

Hint:

To find the threshold frequency, always convert the work function from eV to joules before using \( f = \frac{E}{h} \).

Answer 1

The Correct Option is A

The threshold frequency \( f_0 \) for photoelectric emission is calculated using the relationship between work function (ϕ) and Planck's constant (h):

\( ϕ = h \cdot f_0 \)

The provided values are:

• Work function \( ϕ = 2 \, \text{eV} = 3.2 \times 10^{-19} \, \text{J} \)
• Planck's constant \( h = 6.63 \times 10^{-34} \, \text{Js} \)

Substituting these into the equation yields:

\( 3.2 \times 10^{-19} = 6.63 \times 10^{-34} \times f_0 \)

Solving for \( f_0 \):

\( f_0 = \frac{3.2 \times 10^{-19}}{6.63 \times 10^{-34}} \)

\( f_0 = 4.8 \times 10^{14} \, \text{Hz} \)

The threshold frequency for photoelectric emission is therefore \( 4.8 \times 10^{14} \, \text{Hz} \).