Light of wavelength of 300 nm illuminates a metal with a work function of 1 eV.

Question

Light of wavelength of 300 nm illuminates a metal with a work function of 1 eV. What is the dc Broglie wavelength of the resulting photoelectrons (only one significant number is needed)? (b) Evaluate the commutator for the operators hat x and hat p_x. (c) Is the function, f(x) = x^3 e^-cx^2 - x. even, odd, or neither? (d) For a particle confined to a two-dimensional box, is (x^2 y) equal to (x^2)(y)? Show all math. (e) What is the energy and degeneracy of the first-excited level of a particle confined to a three-dimensional cube of length a?

Explanation / Answer

a)

energy of the emitted photon

Ep = E - W

= ( h c / lamda ) - W

= ( 1240 eV.nm / 300 nm ) - (1 eV )

= 3.133 eV.

wavelength of emitted photon is

lamda ' = hc / Ep

   = ( 1240 eV. nm ) / ( 3.133 eV )

= 395.7 nm

Related Questions

Navigate

• Browse (All)
• Browse L
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.