Question:
In a photoelectric effect experiment the threshold wavelength of the light is 380 nm. If the wavelentgh of incident light is 260 nm, the maximum kinetic energy of emitted electrons will be:
Given E (in eV) = 1237/\(\lambda\) (in nm)
In a photoelectric effect experiment the threshold wavelength of the light is 380 nm. If the wavelentgh of incident light is 260 nm, the maximum kinetic energy of emitted electrons will be:
Given E (in eV) = 1237/\(\lambda\) (in nm)
Given E (in eV) = 1237/\(\lambda\) (in nm)
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\(K_{max}=E_{photon}-\phi\), with both energies from \(E=1237/\lambda\).
Updated On: Nov 5, 2026
- 4.5 eV
- 15.1 eV
- 3.0 eV
- 1.5 eV
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The Correct Option is D
Solution and Explanation
Photon energy: \(E=\dfrac{1237}{260}\approx4.76\) eV.
Work function: \(\phi=\dfrac{1237}{380}\approx3.26\) eV.
\(K_{max}=E-\phi=4.76-3.26\approx1.5\) eV.
Answer: (4) 1.5 eV
Work function: \(\phi=\dfrac{1237}{380}\approx3.26\) eV.
\(K_{max}=E-\phi=4.76-3.26\approx1.5\) eV.
Answer: (4) 1.5 eV
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