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creativepinky
Apr26-10, 03:40 AM
1. The problem statement, all variables and given/known data

In a photoelectric-effect experiment, light of wavelength (x) nm is incident on a sample of sodium. The work function of sodium is (y) J. Calculate the stopping potential required to just stop all electrons from reaching the anode.

2. Relevant equations

To work out frequency, I've used taken c= λ ×f, therefore f= c/λ
which can now be put into:

1/2 m_e v_max^2=hf- ϕ

(which can then be used in the equation v_max=eV_stop to work out the stopping potential.)

3. The attempt at a solution

My question is do you take the value for the electron rest mass (m_e) as 9.109 x 10^-31 to calculate this or do you need to multiply this by the number of electrons that are within sodium (11) and then use this for the value of m_e?

Many thanks in anticipation.

zachzach
Apr26-10, 04:10 AM
1. The problem statement, all variables and given/known data

In a photoelectric-effect experiment, light of wavelength (x) nm is incident on a sample of sodium. The work function of sodium is (y) J. Calculate the stopping potential required to just stop all electrons from reaching the anode.

2. Relevant equations

To work out frequency, I've used taken c= λ ×f, therefore f= c/λ
which can now be put into:

1/2 m_e v_max^2=hf- ϕ

(which can then be used in the equation v_max=eV_stop to work out the stopping potential.)

3. The attempt at a solution

My question is do you take the value for the electron rest mass (m_e) as 9.109 x 10^-31 to calculate this or do you need to multiply this by the number of electrons that are within sodium (11) and then use this for the value of m_e?

Many thanks in anticipation.

You mean KE_{max} = eV_{stop} right? If you multiplied by 11 wouldn't that mean the individual photon released 11 electrons from the atom? Do you even need to use m_e ?

dirtyhippy
Apr26-10, 04:32 AM
K.E max =eVstop = h(f-f0)=hc/λ - y (Work function)

where f is freq, f0 is cutoff freq

the work function (y) can be related
f0 = y/h = c/λ0


hope that helps