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3 years ago

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Electron kinetic energies are often measured in units of electron-volts (1 eV 1.6 x 10-19 J), which is the kinetic energy of an

electron that is accelerated through a 1 volt potential. When an aluminum plate is irradiated with UV light of 253.5 nm wavelength the ejected electrons are observed to have an average kinetic energy of about 0.8 eV. Use these results to determine the electron binding energy (or "work function") o of aluminum (in eV units).

1 answer:

PolarNik [594]3 years ago

5 0

Answer:

4.1 eV

Explanation:

Kinetic energy, K = 0.8 eV = 0.8 x 1.6 x 10^-19 J = 1.28 x 10^-19 J

wavelength, λ = 253.5 nm = 253.5 x 10^-9 m

According to the Einstein energy equation

$E = W_{o}+K$

Where, E be the energy incident, Wo is the work function and K is the kinetic energy.

h = 6.634 x 10^-34 Js

c = 3 x 10^8 m/s

$E=\frac{hc}{\lambda }=\frac{6.634 \times 10^{-34} \times 3 \times 10^{8}}{253.5\times 10^{-9}}=7.85 \times 10^{-19} J$

So, the work function, Wo = E - K

Wo = 7.85 x 10^-19 - 1.28 x 10^-19

Wo = 6.57 x 10^-19 J

Wo = 4.1 eV

Thus, the work function of the metal is 4.1 eV.

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on the x axis : (Ms x Us) + (Mp x Up) = (Ms x Vx) + (Mp x Vx) ...equation 1

on the y axis : (Ms x Us) + (Mp x Up) = (Ms x Vy) + (Mp x Vy) ...equation 2

(0.25 x 25 cos 30) + 0 = (0.25 +0.015) Vx

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therefore the extra distance traveled would be the change in distance = 64.73 -27.6 = 37.13 m

5 0

3 years ago