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###### Interpret energy level diagrams

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10 months ago

Looking at the mercury atom model diagram showing different energy levels, could anyone help me answer this question?

If a source is found to emit an ultraviolet light of wavelength 420nm.

Could this light be emitted by the mercury atom?

What formula do I need to answer this or how do I explain my answer?

My teacher said they don't know how to solve it either, so I feel stuck and hope it won't come up in a test.

If a source is found to emit an ultraviolet light of wavelength 420nm.

Could this light be emitted by the mercury atom?

What formula do I need to answer this or how do I explain my answer?

My teacher said they don't know how to solve it either, so I feel stuck and hope it won't come up in a test.

Reply 1

10 months ago

Original post by Aufbau

Looking at the mercury atom model diagram showing different energy levels, could anyone help me answer this question?

If a source is found to emit an ultraviolet light of wavelength 420nm.

Could this light be emitted by the mercury atom?

What formula do I need to answer this or how do I explain my answer?

My teacher said they don't know how to solve it either, so I feel stuck and hope it won't come up in a test.

If a source is found to emit an ultraviolet light of wavelength 420nm.

Could this light be emitted by the mercury atom?

What formula do I need to answer this or how do I explain my answer?

My teacher said they don't know how to solve it either, so I feel stuck and hope it won't come up in a test.

If you could attach a copy of the diagram that has been provided with the question that would aid with clarity.

Is this not related to finding if the mercury atom can emit photons at the same energy as ultraviolet light with the stated wavelength?

Original post by Joseph McMahon

If you could attach a copy of the diagram that has been provided with the question that would aid with clarity.

Is this not related to finding if the mercury atom can emit photons at the same energy as ultraviolet light with the stated wavelength?

Is this not related to finding if the mercury atom can emit photons at the same energy as ultraviolet light with the stated wavelength?

Hi, I know Ultraviolet light can be anything in the range between 100 - 400nm. But not sure how to calculate this, I calculated the energy differences between energy levels needed to emit a photon and used this to calculate frequencies. but feel stuck when calculating wavelength = speed of light / frequency.

Not getting answers anywhere in the region of 420nm and feel I am completely off track here.

Never imagine a yes or no question would be so complicated to justify.

Reply 4

10 months ago

Original post by Aufbau

Hi, I know Ultraviolet light can be anything in the range between 100 - 400nm. But not sure how to calculate this, I calculated the energy differences between energy levels needed to emit a photon and used this to calculate frequencies. but feel stuck when calculating wavelength = speed of light / frequency.

Not getting answers anywhere in the region of 420nm and feel I am completely off track here.

Never imagine a yes or no question would be so complicated to justify.

Not getting answers anywhere in the region of 420nm and feel I am completely off track here.

Never imagine a yes or no question would be so complicated to justify.

Just to be sure: when calculating energy differences you have converted the eV unit to Joules? (unless you have a formulation that allows you to use eV).

Reply 5

10 months ago

Original post by Aufbau

Looking at the mercury atom model diagram showing different energy levels, could anyone help me answer this question?

If a source is found to emit an ultraviolet light of wavelength 420nm.

Could this light be emitted by the mercury atom?

What formula do I need to answer this or how do I explain my answer?

My teacher said they don't know how to solve it either, so I feel stuck and hope it won't come up in a test.

If a source is found to emit an ultraviolet light of wavelength 420nm.

Could this light be emitted by the mercury atom?

What formula do I need to answer this or how do I explain my answer?

My teacher said they don't know how to solve it either, so I feel stuck and hope it won't come up in a test.

The emitted light is electromagnetic. So what is the speed of this emitted light?

From GCSE, you should know that v = fÎ». Can you calculate f and how might you use it to calculate the energy of the wave? Can you convert the energy of the wave to eV?

Now try calculating some of the energy differences between the energy levels in the mercury atom and see how they compare to the calculated energy?

Original post by Joseph McMahon

Just to be sure: when calculating energy differences you have converted the eV unit to Joules? (unless you have a formulation that allows you to use eV).

Yes, I've done this. (1 eV = 1.60 e-19 J)

Reply 7

10 months ago

Original post by TypicalNerd

The emitted light is electromagnetic. So what is the speed of this emitted light?

From GCSE, you should know that v = fÎ». Can you calculate f and how might you use it to calculate the energy of the wave? Can you convert the energy of the wave to eV?

Now try calculating some of the energy differences between the energy levels in the mercury atom and see how they compare to the calculated energy?

From GCSE, you should know that v = fÎ». Can you calculate f and how might you use it to calculate the energy of the wave? Can you convert the energy of the wave to eV?

Now try calculating some of the energy differences between the energy levels in the mercury atom and see how they compare to the calculated energy?

I would agree with this approach: just convert 420nm to eV and see what results come out of the energy differences based on the eV values. Simpler and more efficient. Then if none come out equal to the 420nm eV value you have your answer.

Reply 8

10 months ago

Original post by Joseph McMahon

I would agree with this approach: just convert 420nm to eV and see what results come out of the energy differences based on the eV values. Simpler and more efficient. Then if none come out equal to the 420nm eV value you have your answer.

When I did A level physics (last year), the approach I explained was the approach expected.

Edit: Misread and misinterpreted your response. Iâ€™m tired lol

(edited 10 months ago)

I actually never had this in GCSE.

Anyway, I work out energy say if an electron moves from level g down to b.

E = E2 - E1

= -2.48 - (-5.74)

= 3.26 ev (1eV = 1.60 e-19 J)

Therefore 3.26ev x 1.60 x 10^-19 J

=5.216 x 10^19 J

Using this to work out frequency I get

f = E/h (plancks constant)

f = 5.216 x 10^-19 J/ 6.63 x 10^-34 Js

f = 7.867 x 10^-54 Hz

I use this to work out the wavelength

Î» = c/f

Î» = 3 x 10^8 / 7.867 x 10^-54

Î» = 3.81 x 10^-47

Where I get stuck is using the relationship between energy and wavelength to convert the energy difference in the first equation to a wavelength.

Î”E = hc/Î»

Anyway, I work out energy say if an electron moves from level g down to b.

E = E2 - E1

= -2.48 - (-5.74)

= 3.26 ev (1eV = 1.60 e-19 J)

Therefore 3.26ev x 1.60 x 10^-19 J

=5.216 x 10^19 J

Using this to work out frequency I get

f = E/h (plancks constant)

f = 5.216 x 10^-19 J/ 6.63 x 10^-34 Js

f = 7.867 x 10^-54 Hz

I use this to work out the wavelength

Î» = c/f

Î» = 3 x 10^8 / 7.867 x 10^-54

Î» = 3.81 x 10^-47

Where I get stuck is using the relationship between energy and wavelength to convert the energy difference in the first equation to a wavelength.

Î”E = hc/Î»

Original post by TypicalNerd

When I did A level physics (last year), the approach I explained was the approach expected.

Edit: Misread and misinterpreted your response. Iâ€™m tired lol

Edit: Misread and misinterpreted your response. Iâ€™m tired lol

I greatly Appreciate any help, totally understand, maybe not the best way to spend a Saturday night. Hope you get some rest

Reply 11

10 months ago

Original post by Aufbau

I actually never had this in GCSE.

Anyway, I work out energy say if an electron moves from level g down to b.

E = E2 - E1

= -2.48 - (-5.74)

= 3.26 ev (1eV = 1.60 e-19 J)

Therefore 3.26ev x 1.60 x 10^-19 J

=5.216 x 10^19 J

Using this to work out frequency I get

f = E/h (plancks constant)

f = 5.216 x 10^-19 J/ 6.63 x 10^-34 Js

f = 7.867 x 10^-54 Hz

I use this to work out the wavelength

Î» = c/f

Î» = 3 x 10^8 / 7.867 x 10^-54

Î» = 3.81 x 10^-47

Where I get stuck is using the relationship between energy and wavelength to convert the energy difference in the first equation to a wavelength.

Î”E = hc/Î»

Anyway, I work out energy say if an electron moves from level g down to b.

E = E2 - E1

= -2.48 - (-5.74)

= 3.26 ev (1eV = 1.60 e-19 J)

Therefore 3.26ev x 1.60 x 10^-19 J

=5.216 x 10^19 J

Using this to work out frequency I get

f = E/h (plancks constant)

f = 5.216 x 10^-19 J/ 6.63 x 10^-34 Js

f = 7.867 x 10^-54 Hz

I use this to work out the wavelength

Î» = c/f

Î» = 3 x 10^8 / 7.867 x 10^-54

Î» = 3.81 x 10^-47

Where I get stuck is using the relationship between energy and wavelength to convert the energy difference in the first equation to a wavelength.

Î”E = hc/Î»

Oh no, this isnâ€™t GCSE level. The equation v = fÎ» is, however, and you are expected to know it at A level.

Not quite. But itâ€™s good to see youâ€™ve attempted to use some correct formulae.

You are told the wavelength is 420 nm. How might you use this, and the speed of the wave to find f? Once you have f, now try finding E using E = hf and follow the approach outlined in my first post.

Okay I've just read some BBC bitesize to get up to speed on the GCSE side of things (I am an access student btw, so we get 2 weeks to learn a Module and than cram it all into a 2 hour test).

So I am first going to convert the wavelength into meters (Î» = 420 nm Ã— 10^-9 m/nm = 4.2 Ã— 10^-7 m)

f = v/Î» = 3 x 10^8 m s/ 4.2 x 10^-7 m

f = 7.143 x 10^14

------

E = hf

E = (6.63x 10^-34) x (7.143 x 10^14)

= 4.736x10^19

And I think I've done something wrong, as this doesn't match any of the values.

So I am first going to convert the wavelength into meters (Î» = 420 nm Ã— 10^-9 m/nm = 4.2 Ã— 10^-7 m)

f = v/Î» = 3 x 10^8 m s/ 4.2 x 10^-7 m

f = 7.143 x 10^14

------

E = hf

E = (6.63x 10^-34) x (7.143 x 10^14)

= 4.736x10^19

And I think I've done something wrong, as this doesn't match any of the values.

Reply 13

10 months ago

Original post by Aufbau

Okay I've just read some BBC bitesize to get up to speed on the GCSE side of things (I am an access student btw, so we get 2 weeks to learn a Module and than cram it all into a 2 hour test).

So I am first going to convert the wavelength into meters (Î» = 420 nm Ã— 10^-9 m/nm = 4.2 Ã— 10^-7 m)

f = v/Î» = 3 x 10^8 m s/ 4.2 x 10^-7 m

f = 7.143 x 10^14

------

E = hf

E = (6.63x 10^-34) x (7.143 x 10^14)

= 4.736x10^19

And I think I've done something wrong, as this doesn't match any of the values.

So I am first going to convert the wavelength into meters (Î» = 420 nm Ã— 10^-9 m/nm = 4.2 Ã— 10^-7 m)

f = v/Î» = 3 x 10^8 m s/ 4.2 x 10^-7 m

f = 7.143 x 10^14

------

E = hf

E = (6.63x 10^-34) x (7.143 x 10^14)

= 4.736x10^19

And I think I've done something wrong, as this doesn't match any of the values.

E should be 4.736 x 10^-19 J (I assume that is a typo, since all your working and digits look correct to me). Notice how the energies on the diagram are given in eV? Now convert E into eV.

Then calculate the differences in the energy levels on the diagram and compare them to your calculated value of E.

(edited 10 months ago)

4.736 x 10^-19 / 1.602 x 10^-19 J

= 2.96 eV

WOW! Thank you.

Okay, since my value is between energy levels f and e, I would say No, Mercury could not emit a wavelength of 420nm, because Photons are only emitted going down energy levels, and must be a certain value as shown in the calculations above.

= 2.96 eV

WOW! Thank you.

Okay, since my value is between energy levels f and e, I would say No, Mercury could not emit a wavelength of 420nm, because Photons are only emitted going down energy levels, and must be a certain value as shown in the calculations above.

Reply 15

10 months ago

Original post by Aufbau

4.736 x 10^-19 / 1.602 x 10^-19 J

= 2.96 eV

WOW! Thank you.

Okay, since my value is between energy levels f and e, I would say No, Mercury could not emit a wavelength of 420nm, because Photons are only emitted going down energy levels, and must be a certain value as shown in the calculations above.

= 2.96 eV

WOW! Thank you.

Okay, since my value is between energy levels f and e, I would say No, Mercury could not emit a wavelength of 420nm, because Photons are only emitted going down energy levels, and must be a certain value as shown in the calculations above.

You could make reference to energy being quantised (the jargon word for just saying you can't have any in-between values).

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