Question

Calculate the velocity of an electron ejected if 660.0 nm of light is applied to the surface. A wavelength of 795 nm has sufficient energy to eject electrons.

Answer 1

Explanation:

Unless light of sufficient frequency is used, then no electrons are ejected. ... kinetic energy of the electron is the energy of the photon minus the threshold energy. ... This is the same as the above equation re-arranged

Answer 2

$\huge{\underline{\underline{\mathfrak{Answer}}}}$

Photoelectric equation:

hf = Φ + KE

Φ = work function (J) = energy needed to remove an electron

h = Planck’s constant = 6.626 x 10^-34 J • s

f = threshold frequency or higher

KE = maximum kinetic energy of emitted electrons (J)

c = speed of light = 2.998 x 10^8 m/s

mass of electron = 9.109 x 10^-31 kg

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The energy of a photon of 795 nm light will give you the work function Φ .

λ = 795 nm = 7.95 x 10^-7 m

E = (hc/λ) = 2.499 x 10^-19 J

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Find the energy of a photon with wavelength λ = 336.0 nm = 3.360 x 10^-7 m

E = (hc/λ) = 5.912 x 10^-19 J

This is the hf term in the photoelectric equation. [hf = (hc/λ)]

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The kinetic energy is

KE = hf - Φ = (5.912 x 10^-19 J) - (2.499 x 10^-19 J) = 3.413 x 10^-19 J

KE = (1/2)mv² , so v = √(2 KE / m)

v = √[(2 * 3.413 x 10^-19 J) / (9.109 x 10^-31 kg)]

= 8.66 x 10^5 m/s