Quantum Question: Max Kin Energy & Speed of Electron

[DarkKnight911]

Homework Statement

Light with a wavelength 600 nm is directed at a mettalic surface with a work function of 1.6 eV. A retarding potential of 0.5 V is applied. Determine:
a) The maximum kinetic energy, in joules, of an emitted electron before the retarding potential is applied.
b) The maximum speed with which an electron reaches the far side after the retarding potential is applied.

Homework Equations

[/B]
Kinetic Energy = hf - w

The Attempt at a Solution

I used the above equation to figure out "a" but, I do not know where to begin question "b."

 

[ehild]

Welcome to PF!

The energy of the electron is conserved during its flight. Retarding potential means that it travels toward an electrode at negative potential with respect to the metallic surface. How much does the potential energy of the electron change?

 

[DarkKnight911]

Oh I see but can you please do it for me I just want a solid example to follow in the future.

 

[ehild]

Oh I see but can you please do it for me I just want a solid example to follow in the future.

I can not do it for you, the rules of the Forums forbid it. We only can lead you.

Do you know what is the potential energy of a charge q at a point with potential U?

 

[HalfLostAndSearching]

For part b), we can use the equation for kinetic energy to find the maximum speed of the electron. First, we need to find the maximum kinetic energy of the electron before the retarding potential is applied. This can be found using the equation:

Kinetic Energy = hf - w

where h is Planck's constant (6.626 x 10^-34 J*s) and f is the frequency of the light, which can be found using the equation:

f = c/λ

where c is the speed of light (3.00 x 10^8 m/s) and λ is the wavelength of the light (600 nm = 6.00 x 10^-7 m).

So, we have:

f = (3.00 x 10^8 m/s)/ (6.00 x 10^-7 m) = 5.00 x 10^14 Hz

Now, we can plug this into the equation for kinetic energy:

Kinetic Energy = (6.626 x 10^-34 J*s)(5.00 x 10^14 Hz) - (1.6 eV)(1.602 x 10^-19 J/eV) = 4.20 x 10^-19 J

This is the maximum kinetic energy of the electron before the retarding potential is applied.

To find the maximum speed of the electron after the retarding potential is applied, we can use the equation for kinetic energy again:

Kinetic Energy = 1/2 mv^2

where m is the mass of the electron (9.11 x 10^-31 kg) and v is the velocity of the electron.

We already know the value for kinetic energy from part a), so we can plug that in and solve for v:

4.20 x 10^-19 J = 1/2 (9.11 x 10^-31 kg) v^2

Solving for v, we get:

v = 1.40 x 10^6 m/s

So, the maximum speed with which the electron reaches the far side after the retarding potential is applied is 1.40 x 10^6 m/s.