When photons of energy 4.25 eV strike the surface of a metal A, the ejected…
Chemistry · JEE Main · NTA Exams — Atomic Structure
When photons of energy 4.25 eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy TA eV and de Broglie wavelength λA.
The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70 eV is \(\mathrm{T}_{\mathrm{B}}=\left[\mathrm{T}_{\mathrm{A}}-1.50\right] \mathrm{eV}\).
If the de Broglie wavelength of these photoelectrons is \(\lambda_{\mathrm{B}}=2 \lambda_{\mathrm{A}}\), then find the incorrect option.
- the work function of A is 2.25 eV
- the work function of B is 4.20 eV
- \(\mathrm{T}_{\mathrm{A}}=2.00 \mathrm{eV}\)
- \(\mathrm{T}_{\mathrm{B}}=2.75 \mathrm{eV}\)
Answer
(B) the work function of B is 4.20 eV
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