First Part A magnesium surface has a work function of 3.68 eV. Electromagnetic w

Question

First Part

A magnesium surface has a work function of 3.68 eV. Electromagnetic waves with a wavelength of 215 nm strike the surface and eject electrons. What is the maximum kinetic energy of the ejected electrons?

Answer:

    E = hc/? = 6.63x10-34x3.0x108 /215x10-9

                    = 9.25x10-19 J = 5.78 eV

Hence, Kmax = E - w = 5.78 - 3.68= 2.10 eV

Remanining Questions

1) If the intensity of the light striking the surface is increased, what will happen to the maximum kinetic energy of the ejected electrons?

2) If the wavelength of the light striking the surface is increased, what will happen to the maximum kinetic energy of the ejected electrons?

3) If a metal with a greater work function were used, what will happen to the maximum kinetic energy of the ejected electrons?

A) Increase

B) Decrease

C) Same

Explanation / Answer

1) If the intensity of the light striking the surface is increased, what will happen to the maximum kinetic energy of the ejected electrons?

Intensity is independent to Ke of the ejected electrons since intensity depends on the number of photons emitted

Intensity of radiation~ Number of photons emitted

Hence, maximum kinetic energy of the ejected electrons remains same

2) If the wavelength of the light striking the surface is increased, what will happen to the maximum kinetic energy of the ejected electrons?

Here wavelength of the light striking the surface is dependent to Ke of the ejected electrons since E=hf= hc/lamda

Wavelength of the light striking the surface is inversely proportional to Ke

Hence, maximum kinetic energy of the ejected electrons decreases

3) If a metal with a greater work function were used, what will happen to the maximum kinetic energy of the ejected electrons?

Here Kemax= E-work function

The maximum possible kinetic energy occurs when the work function is minimum

Thereby, when work function increases kinetic energy decreases

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