Physics - Got Some Questions (1 Viewer)

[Life'sHard]

 

[idkkdi]

View attachment 32032

what minimum value do they want lol

 

[notme123]

what minimum value do they want lol

min value of the voltage at the largest rotational velocity which makes me also think 0. At that point, the w would be at equilibrium

 

[notme123]

a more interesting q would be for us to predict max rpm but as you can see the voltage decrease isn't exactly linear, in fact theres no correlation at all.
the only relation ship is as rpm increase so does therefore decreases

 

[Life'sHard]

Can someone help explain what's happening here and why it's a jump from 3 to 2. Thanks.

 

[notme123]

View attachment 32639
Can someone help explain what's happening here and why it's a jump from 3 to 2. Thanks.

as fringe spacing increases, so does wavelength. The longest wavelength in the options is 3-->2 which corresponds to the farthest away bright band.

 

[Life'sHard]

as fringe spacing increases, so does wavelength. The smallest wavelength in the options is 3-->2 which corresponds to the farthest away bright band.

Oh I see. This might sound a little stupid lol but what exactly is 3 and 2. Are those shell numbers?

 

[notme123]

Oh I see. This might sound a little stupid lol but what exactly is 3 and 2. Are those shell numbers?

ye balmer series is when a electron transitions down to the second shell

 

[Life'sHard]

ye balmer series is when a electron transitions down to the second shell

Oh ok. The way I counted it was P was the 1st shell so n=1 and then Q was the 5th shell so I had thought it was 5 down to 2. Can you reason why this is incorrect?

 

[notme123]

Oh ok. The way I counted it was P was the 1st shell so n=1 and then Q was the 5th shell so I had thought it was 5 down to 2. Can you reason why this is incorrect?

emission spectra are caused by transition in shells, which emits a photon of a certain wavelength. When you diffract all these diffrent wavelegths, you will get bright fringes on a screen according to . So that means as the wavelength of the diffracted light increases, the angle of diffraction away from the central bright spot also increases. Since Q is furthest away from the central bright band, this means the light at Q has the longest wavelength, meaning the photons also have the least energy. The smallest energy gap between transition states is 3-->2. It would be false to use your method as a transition from 5--> 2 has the largest energy gap out of the options, meaning a shorter wavelength, so it's bright band would appear closer to the central band. Assuming the middle is 1 is also wrong as for an electron to transition from 1-->2 it needs to absorb energy, not emit, so under that assumption there would be no bright spot there, neither at the next ones over.

 

[Life'sHard]

emission spectra are caused by transition in shells, which emits a photon of a certain wavelength. When you diffract all these diffrent wavelegths, you will get bright fringes on a screen according to . So that means as the wavelength of the diffracted light increases, the angle of diffraction away from the central bright spot also increases. Since Q is furthest away from the central bright band, this means the light at Q has the longest wavelength, meaning the photons also have the least energy. The smallest energy gap between transition states is 3-->2, It would be false to use your method as a transition from 5--> 2 has the largest energy gap out of the options, meaning a shorter wavelength, so it's bright band would appear closer to the central band. Assuming the middle is 1 is also wrong as for an electron to transition from 1-->2 it needs to absorb energy, not emit, so under that assumption there would be no bright spot there, neither at the next ones over.

I see I see. So as the fringes move further away from the centre (P), the wavelength of light becomes longer. This results in photons having a smaller amount of energy which is (Q). Therefore considering the difference between the shell no.s the lowest difference is 3-->2.
This diagram confused me tho. Cause from the question I had thought each band represented a shell number (n) which I now know is incorrect.

 

[notme123]

I see I see. So as the fringes move further away from the centre (P), the wavelength of light becomes longer. This results in photons having a smaller amount of energy which is (Q). Therefore considering the difference between the shell no.s the lowest difference is 3-->2.View attachment 32641
This diagram confused me tho. Cause from the question I had thought each band represented a shell number (n) which I now know is incorrect.

it's just like young double slit experiment basically. If you ever look at white light diffracted, purple is always closest to the centre while red is out on the edges, implying light with more energy (purple) are closer to the centre than light with less energy (red)

 

[Life'sHard]

it's just like young double slit experiment basically. If you ever look at white light diffracted, purple is always closest to the centre while red is out on the edges, implying light with more energy (purple) are closer to the centre than light with less energy (red)

Sweet! Tyty.