It requires a photon with a minimum energy of 4.41 ? 10 -19 J to emit electrons

Question

It requires a photon with a minimum energy of 4.41 ? 10-19 J to emit electrons from sodium metal

a)What is the minimum frequency of light necessary to emit electrons from sodium via the photoelectric effect? I got 6.66x10^14 as this answer

I don't understand parts b-d

b)What is the wavelength of this light?
c) If sodium is irradiated with light of 432 nm, what is the maximum possible kinetic energy of the emitted electrons?

d)What is the maximum number of electrons that can be freed by a burst of light (? = 432 nm) whose total energy is 1.29

Explanation / Answer

B)Use c=lambda x nu, rearrange to c/v=lamba(wavelength)
we know

C=3.00x10^8m/s

wavelengh= 3.00x10^8/6.66x10^14

=4.50x10^-7m

divide by 10^-9m= 450 nm

C)

E1 = 6.626E-34 * 3E8 / 432E-9 = 4.601E-19 J/electron

minimum energy of 4.41E-19 J/electron

so maximum possible kinetic energy of the emitted electrons= E-E1

4.601E-19 - 4.41E-19 = 0.191 E-19J/electron

D)1.29 micrometer equals 1.29 x 10^-6J E = hc/w

E = 6.626E-34 * 3E8 / 432E-9 = 4.601E-19 J/electron

1.29E-6 J *(1 electron / 4.601E-19 J) = 2.804 x10^12 electrons

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