Question:
If the work function of a metal is 6.875 eV, its threshold wavelength will be (Take \(c = 3 \times 10^8\) m/s)
If the work function of a metal is 6.875 eV, its threshold wavelength will be (Take \(c = 3 \times 10^8\) m/s)
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Useful constant: \(hc = 1240\) eV·nm = 12400 eV·Å.
Updated On: Apr 23, 2026
- 3600 Å
- 2400 Å
- 1800 Å
- 1200 Å
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept:
Threshold wavelength: \(\lambda_0 = \frac{hc}{W}\). \(h = 6.63 \times 10^{-34}\) J s, \(1\) eV = \(1.6 \times 10^{-19}\) J.
Step 2: Detailed Explanation:
\(W = 6.875 \times 1.6 \times 10^{-19} = 1.1 \times 10^{-18}\) J.
\(\lambda_0 = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{1.1 \times 10^{-18}} = \frac{1.989 \times 10^{-25}}{1.1 \times 10^{-18}} = 1.808 \times 10^{-7}\) m = 1808 Å ≈ 1800 Å.
Alternatively: \(\lambda_0(\text{Å}) = \frac{12400}{W(\text{eV})} = \frac{12400}{6.875} \approx 1804\) Å.
Step 3: Final Answer:
Thus, threshold wavelength = 1800 Å.
Threshold wavelength: \(\lambda_0 = \frac{hc}{W}\). \(h = 6.63 \times 10^{-34}\) J s, \(1\) eV = \(1.6 \times 10^{-19}\) J.
Step 2: Detailed Explanation:
\(W = 6.875 \times 1.6 \times 10^{-19} = 1.1 \times 10^{-18}\) J.
\(\lambda_0 = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{1.1 \times 10^{-18}} = \frac{1.989 \times 10^{-25}}{1.1 \times 10^{-18}} = 1.808 \times 10^{-7}\) m = 1808 Å ≈ 1800 Å.
Alternatively: \(\lambda_0(\text{Å}) = \frac{12400}{W(\text{eV})} = \frac{12400}{6.875} \approx 1804\) Å.
Step 3: Final Answer:
Thus, threshold wavelength = 1800 Å.
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