Physics: Photoelectric effect equation, understanding Planck's Constant?

Start with E = hf ; where E is the energy of the incident photon (J or eV) and f is the frequency (in Hz) of the light being used.

However, we also know that the frequency, f, can be also written as: f= c/lambda; where c is the velocity of light (3 x 10^8 ms-1) and lambda is the wavelength in metres.

Using your terminology, E = W + KE or hf = W + KE.

Where W is the work Function and KE is the kinetic energy.

If you decide to work in electron volts(or eV) then all three terms should be in eV. If however, you decided to work in joules all three terms should be in Joules.

If I understand you correctly, what you need to plot is KE in eV on the 'y' axis and frequency in Hz on the 'x' axis

This means that to start with we need to re-arrange our expression to be : KE = hf - W, just like you said! Since you appear to have written KEmax = eVs ..... I assume you want to work in electron Volts. This means that the other two terms must also be in electron volts (often, the work function, W, is espressed in eV anyway!)

CONVERTING THE FIRT LINE: Red - 650 - 8.2E+05

Step 1: convert wavelength to frequency

First, you need to convert the wavelength give in nm into frequency using f = c/lambda.

So, 650 nm gives: f = 3x10^8/650 x 10^-9 = 4.615 x 10^14 Hz

This means that our line now looks like: Red - 4.62 x 10^14 - 8.2E+05

Step 2: Convert velocity(max) to KEmax

Secondly, we need to convert the ejection velocity into KE, for this we use KE = 0.5 m x v^2

Where v in this case is: 8.2 x 10^5 ms-1.

It's a little bit perverse of your teacher not to give you the mass of the electron; but, since you are given the e/m ratio is 1.76 x 10^11 Ckg-1; and the electronic charge, e = 1.6 x 10^-19 then we can evaluate m as follows:

e/m = 1.76 x 10^11, therefore, 1.6 x 10^-19/m = 1.76 x 10^11 giving m as 1.6 x 106-19/1.76 x 10^11

m =9.09 x 10^-31 kg or 9.1 x 10^-31kg

Hence the KE is 0.5 x 9.1 x 10^-31 x (8.2 x 10^5)^2 = 3.059 x 10^-19 J

Step 3 convert KE to Joules

Converting to eV we get: KE = 3.059 x 10^-19 / 1.6 x 10^-19 = 1.912 eV

So, the first line finally looks like this

Colour f(Hz) KE (ev)

Red - 4.62 x 10^14 - 1.91

I have done the also the green and the violet

Colour f(Hz) KE(ev)

Green - 5.66 x 10^14 - 2.35

Violet - 7.143 x 10^14 - 2.84

You need to fill in the others to be be able to convince your teacher you know what you are doing!

If you need help e-mail me or post an additional message requesting help.

Your graph of KE (ev) measured on the 'y' axis and frequency of the 'x' axis comes from:

KE = hf - W. This is a straight line of the form y = mx +c

KE = h f + W

y = m x + c

Where: y is the KE; x is the frequency , f; m is the gradient of the straight line (h Planks constant) and c is where the line intercepts the y axis at x =0 and this will read off the work function of the metal in eV.

The 'y' axis (KE) will be in eV ranging from 1.91 to 2.84, the 'x' axis will be in Hz x 10^14 ranging from 4.62 - 7.1. The line will slope downwards from left to right.

The gradient, m, will give Planks constant. Taking the two limits (red and violet).

gradient, m = change in y/ change in x

m = change in KE/change in frequency

m = (1.91-2.84)/(4.62-7.14)x 10^14 = -0.93/-2.52 x 10^14 = 3.69 x 10^-14

to get planks constant out we need to multiply by 1.6 x 10^-19 since we did not convert the energy to eV before.

m = planks contstant comes out as 3.69 x 10^-14 x 1.6 x 10^-19 = 5.9 x 10^-34 Js

This is in reasonable agreement to the theoretical value given experimental errors involved in any measuremets made.

Hope this helps.

No need for BA or points.

Photoelectric Effect Equation

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RE:

Physics: Photoelectric effect equation, understanding Planck's Constant?

Charge of the electron = 1.6 x 10-19 C

Charge to mass ratio for the electron = 1.76 x 1011 C/kg

Photoelectric Effect Data:

Light color - Wavelength (nm) - Electron Velocity (m/s)

Red - 650 - 8.2E+05

Orange - 600 - 8.5E+05

Yellow - 580 - 8.7E+05

Green - 530 - 9.1E+05

Cyan -...

sure, i have time to walk you through the concept of photoelectric effect

the law said

when the energy of the [incident light] is greater than the [work function] of the metal, [electron is emitted]

we have 3 things here (1) incident light (2) work function (3) emitted electron, i'll put that into an equation

[ incident light photon energy ] = [ metal work function energy ] + [ KE of emitted electron ]

(1) incident light photon energy

for energy unit let used eV

planck said photon energy = (h)(f) = (10^9)(h)(c)(w)

10^9 conversion for wavelength nm, red light 650nm

photon energy = (10^9)(4.135 667E-15)(299 792 458)(650)

(2) metal work function

consult the book

lets say we used cesium material which they used this material to turn on and off the highway street light

cesium work function = 2.14 eV

(3) ke of emitted electron

ke = (1/2)(m)(v^2)

lets see if you are given data for this,

i know the mass of electron m=9.11E-31kg but there is no velocity (v), but don't worry, lets used another equation

ke = (1/2)(m)(v^2) = (Q)(Vo)

when they design the photoelectric circuit, they have a battery supplying a volatge across the cesium material that gives Vo= 5 volts dc, but you don't have this data, you have Q=1.6E-19C, now i'm start worrying and have to cheat a little bit by assuming the stopping voltage Vo is 5 volts dc,

so ke = (1.6E-19)(5) or lets assumed when the electron is emitted it has a velocity of 3E+5 m/s,

so ke =(1/2)(9.11E-31)(3E+5)^2

ops! sorry, my mistake, if i read you correctly, you want to find out the ke of electron flying out from the metal

okay, now we have a complete equation

[ incident light photon energy ] = [ metal work function energy ] + [ KE of emitted electron ]

[ (10^9)(4.135 667E-15)(299 792 458)(650) ] = [ 2.14 ] + [ KE ]

solve for KE <= ans

explanation is too messy and too long, sorry

energy work function value of metal is from CRC handbook