A 2.50-W beam of light of wavelength 124 nm falls on a metal surface. You observe that the maximum kinetic energy of the ejected electrons is 4.16 eV. Assume that each photon in the beam ejects a photoelectron. (a) What is the work function (in electron volts) of this metal

Respuesta :

Answer:

φ = 13.43 x 10⁻¹⁹ J = 8.4 eV

Explanation:

Using the Einstein's Photoelectric equation:

Energy of Photon = Work Function + Kinetic Energy of Electron

[tex]\frac{hc}{\lambda} = \phi + K.E[/tex]

where,

h = Plank's Constant = 6.625 x 10⁻³⁴ J.s

c = speed of light = 3 x 10⁸ m/s

λ = wavelength = 124 nm = 1.24 x 10⁻⁷ m

φ = work function = ?

K.E = Kinetic Energy of Electrons = (4.16 eV)([tex]\frac{1.6\ x\ 10^{-19}\ J}{1\ eV}[/tex]) = 2.6 x 10⁻¹⁹ J

Therefore,

[tex]\frac{(6.625\ x\ 10^{-34}\ J.s)(3\ x\ 10^8\ m/s)}{1.24\ x\ 10^{-7}\ m} = \phi + 2.6\ x\ 10^{-19}\\\\\phi=16.03\ x\ 10^{-19}\ J - 2.6\ x\ 10^{-19}\ J[/tex]

φ = 13.43 x 10⁻¹⁹ J = 8.4 eV