When photons of energy 4.25 eV strike the surface of a metal A, the ejected…

Chemistry · JEE Main · NTA ExamsAtomic Structure

When photons of energy 4.25 eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy TeV 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.

  1. the work function of A is 2.25 eV
  2. the work function of B is 4.20 eV
  3. \(\mathrm{T}_{\mathrm{A}}=2.00 \mathrm{eV}\)
  4. \(\mathrm{T}_{\mathrm{B}}=2.75 \mathrm{eV}\)

Answer

(B) the work function of B is 4.20 eV

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