# Maximum Frequency of an emitted photonWatch

Thread starter 7 years ago
#1
I'm following the June 2009 Unit 1 Paper, and dont understand the mark scheme for Q2

In the previous question, I worked out 9ev as being 1.44 × 10-18

QUESTION

Calculate the maximum frequency of the photon emitted when the 9.0 ev
electron collides with an atom.

MARK SCHEME

E = 1.44 × 10-18 - 1.6 × 10-19 = 1.28 × 10-18

6.63 × 10-34 × f = 1.28 × 10-18

It's the first line I don't get.. where does the 1.6 x 10-19 (Charge of proton/election) come from? im sure there's something im missing
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7 years ago
#2
The 1.6 x 10-19 is the actually charge of the electron. It is a known value that is usually given or you learn yourself.
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Thread starter 7 years ago
#3
(Original post by Wick3d)
The 1.6 x 10-19 is the actually charge of the electron. It is a known value that is usually given or you learn yourself.
I know, but why is the 1.6 x 10-19 taken away from the energy value I worked out?

Why couldn't the 1.44 x 10-18 be used in the the E = hf ?
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7 years ago
#4
the answer should be 2.17 x 10 to the power 15

the mark scheme seems slighly dubious.
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7 years ago
#5
Ok, this is how I see it:

In the question it states that an electron is giving off a photon of energy.
The electron must emit the extra energy but must also retain the amount of energy for one electron,
the energy of one electron has to be subtracted from the energy that the excited electron has before the emission of the photon.
Knowing this:
9.0eV = 1.44*10-18J
Energy of an electron = 1.6*10-19
1.44*10-18 - 1.6*10-19 = 1.28*10-18
6.63*10-34 * f = 1.28*10-18
Therefore...
f = 1.28*10-18 / 6.63*10-34 = 1.9*10 15Hz
So...
The maximum frequency of the photon emitted = 1.9*10 15Hz
Hope this makes the solution clearer to understand.
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6 years ago
#6
(Original post by gouldie70)
I'm following the June 2009 Unit 1 Paper, and dont understand the mark scheme for Q2

In the previous question, I worked out 9ev as being 1.44 × 10-18

QUESTION

Calculate the maximum frequency of the photon emitted when the 9.0 ev
electron collides with an atom.

MARK SCHEME

E = 1.44 × 10-18 - 1.6 × 10-19 = 1.28 × 10-18

6.63 × 10-34 × f = 1.28 × 10-18

It's the first line I don't get.. where does the 1.6 x 10-19 (Charge of proton/election) come from? im sure there's something im missing
The first part says that 9eV before collision, second part says 1 eV after collision therefore only 8eV is used.
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5 years ago
#7
I see what it's done, it's done it the most confusing way but it is correct.
What it's done is it's converted the eV into Joules straight away, and it's worked out what the Energy in Joules that it would have been and then subtracted them away. So it's done (1.44x10^-19) (9eV) - (1.6x10^-19) (1eV) which leaves you with 1.28x10^-18 (8eV)

Then you use your E = hf to work it out

1.28x10^-18 = (6.63x10^-34) x f

Rearrange the formula and Hey Presto, you get the answer in the booklet, 1.9(3)x10^15 Hz

As a Sixth Former it messes with my head (crycry!) but I hope that helps anyone else who finds this question confusing!
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