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a.) If sodium is irradiated with light of 439 nm, what is themaximum possible kinetic energy of the emitted electrons?

b.) What is the maximum number of electrons that can be freedby a burst of light whose total energy is 1.00μJ?

Respuesta :

Answer:

A. 4.528 x 10-19 J

B. 2.208 x 1012 electrons

Explanation:

A. The energy of a photon can be got from Plancks equation;

E = hc/wavelength = hv

Where h = heinsbergs constant = 6.626 x 10-34J

c = speed of light = 3 x 108m/s

Wavelength = 439nm = 4.39 x 10-7 m

E = (6.626 x10-34 * 3 x 108)/ 4.39 x 10-7

= 4.528 x 10-19 J

Sodium metal requires a photon with a minimum energy of 4.41 x 10-19 J to emit electrons.

The maximum kinetic energy for each freed electron will be;

4.528 x 10-19 - 4.41 x 10-19

= 1.18 x 10-20J

B. Each photon can free at least 1 electron, therefore;

1 photon = 1 electron

1 x 10-6/4.528 x 20-19

= 2.208 x1012 photons

= 2.208 x 1012 electrons

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