Photons having energy equivalent to the binding energy of the 4th state of a He⁺ ion are used on a metal surface with a work function of 1.4 eV. If electrons are further accelerated through a potential difference of 4V, the minimum de-Broglie wavelength associated with the electron is:
Correct answer: C. 5 Å
- A. 1.1 Å
- B. 9.15 Å
- C. 5 Å
- D. 11 Å
Explanation
The binding energy of the 4th state of He⁺ is E = 13.6(Z²/n²) = 3.4 eV. The kinetic energy of the ejected electron is 3.4 - 1.4 = 2.0 eV. After acceleration, the final KE is 2.0 + 4.0 = 6.0 eV. The de-Broglie wavelength is λ = h/√(2mKE), which calculates to approximately 5 Å.
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