Question

The work function of a material is $6.6 \text{ eV}$. Then, the threshold wavelength of the metal is approximately (Take $h = 6.6 \times 10^{-34} \text{J.s}$)}

• A. $108 \text{ nm}$
• B. $188 \text{ nm}$
• C. $208 \text{ nm}$
• D. $228 \text{ nm}$
• E. $250 \text{ nm}$

Hint:

For rapid calculations in exams, use $\lambda_0 (\text{in nm}) \approx \frac{1240}{E (\text{in eV})}$.
Here, $\frac{1240}{6.6} \approx 187.8 \text{ nm}$, which is closest to $188 \text{ nm}$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The work function is the minimum energy required to eject an electron from a metal surface. The threshold wavelength is the maximum wavelength of incident light that corresponds to exactly this energy.
Step 2: Key Formula or Approach:
Work function equation: \(W = h\nu_0 = \frac{hc}{\lambda_0}\).
Rearranging for threshold wavelength: \(\lambda_0 = \frac{hc}{W}\).
Convert work function from eV to Joules: \(1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}\).
Step 3: Detailed Explanation:
Given values:
\(W = 6.6 \text{ eV} = 6.6 \times 1.6 \times 10^{-19} \text{ J}\)
\(h = 6.6 \times 10^{-34} \text{ J s}\)
\(c = 3 \times 10^8 \text{ m/s}\)
Substitute into the formula:
\[ \lambda_0 = \frac{(6.6 \times 10^{-34})(3 \times 10^8)}{6.6 \times 1.6 \times 10^{-19}} \] Cancel the 6.6 from numerator and denominator:
\[ \lambda_0 = \frac{3 \times 10^{-26}}{1.6 \times 10^{-19}} \] \[ \lambda_0 = \frac{3}{1.6} \times 10^{-7} \text{ m} \] \[ \lambda_0 = 1.875 \times 10^{-7} \text{ m} \] To convert meters to nanometers (nm), multiply by \(10^9\):
\[ \lambda_0 = 1.875 \times 10^{-7} \times 10^9 = 187.5 \text{ nm} \] This is approximately 188 nm.
Step 4: Final Answer:
The threshold wavelength is approximately 188 nm.