magnesium metal foil was illuminated by a pulse of laser beam with a wavelength of 306 nm. A total number of 6 x 10³ ejected electrons were generated. Each electron has an de Broglie wavelength of 2 nm. (a) Calculate the work function (express the answer in eV) (b) Calculate the threshold frequency (express the answer in nm).

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answer parts a and b

magnesium metal foil was illuminated by a pulse of laser beam with a wavelength of 306 nm. A
total number of 6 × 10³ ejected electrons were generated. Each electron has an de Broglie
wavelength of 2 nm.
(a) Calculate the work function (express the answer in eV)
(b) Calculate the threshold frequency (express the answer in nm).
Transcribed Image Text:magnesium metal foil was illuminated by a pulse of laser beam with a wavelength of 306 nm. A total number of 6 × 10³ ejected electrons were generated. Each electron has an de Broglie wavelength of 2 nm. (a) Calculate the work function (express the answer in eV) (b) Calculate the threshold frequency (express the answer in nm).
Expert Solution
Step 1: de Broglie relation and Photoelectric effect formula

The de Broglie wavelength (lambda) of a particle with momentum p is given by:

lambda equals h over p

where h is Planck's constant (6.626 cross times 10 to the power of negative 34 end exponent J·s) and p is the momentum of the particle.

The photoelectric effect relates the energy of a photon to the work function (straight capital phi) and the kinetic energy of the emitted electron (K. E.):

E subscript text photon end text end subscript equals straight capital phi plus K. E.

Given that E subscript text photon end text end subscript equals h nu, where h is Planck's constant and nu is the frequency of the incident light.

The kinetic energy (K. E.) of the emitted electron can be expressed in terms of its velocity (v) as:

K. E. equals 1 half m v squared

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