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 ($\phi_0$) of a metal is the minimum energy required to eject an electron from its surface. The corresponding maximum wavelength is the threshold wavelength ($\lambda_0$).
Key Formula or Approach:
\[ \phi_0 = \frac{hc}{\lambda_0} \implies \lambda_0 = \frac{hc}{\phi_0} \]
Note: $1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}$.

Step 2: Detailed Explanation:
Given:
Work function $\phi_0 = 6.6 \text{ eV} = 6.6 \times 1.6 \times 10^{-19} \text{ J}$.
$h = 6.6 \times 10^{-34} \text{ J.s}$.
$c = 3 \times 10^8 \text{ m/s}$.
Substitute the values:
\[ \lambda_0 = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{6.6 \times 1.6 \times 10^{-19}} \]
The $6.6$ cancels out 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 to nm ($10^{-9} \text{ m}$):
\[ \lambda_0 = 187.5 \text{ nm} \approx 188 \text{ nm} \]

Step 3: Final Answer:
The threshold wavelength is approximately $188 \text{ nm}$.