The Student Room Group
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>

Plotting straight line graphs from physics equations question

I am having trouble with a question which goes as follows:
- What would you plot on the x and y axes of a graph to obtain a straight line for each of the following given that you can only directly measure the 2 variables listed. For each one, what is the gradient and y-axis intercept

- The equation in question goes as follows: E=(hc)/ λ
- The variables you can measure are E and λ.

Any help would be great
Original post by Alexokay
I am having trouble with a question which goes as follows:
- What would you plot on the x and y axes of a graph to obtain a straight line for each of the following given that you can only directly measure the 2 variables listed. For each one, what is the gradient and y-axis intercept

- The equation in question goes as follows: E=(hc)/ λ
- The variables you can measure are E and λ.

Any help would be great

A graph of E against 1/ λ.

Gradient will be hc, and c is known usually rounded to 3×10^8 m/s. h will be 6.63×10^-34
Reply 2
Avatar for Alexokay
Alexokay
OP
11
Original post by Knightwing98124
A graph of E against 1/ λ.Than

Gradient will be hc, and c is known usually rounded to 3×10^8 m/s. h will be 6.63×10^-34

Thanks, I guessed the gradient although I am struggling to understand why the x axis is 1/ λ???
Original post by Alexokay
Thanks, I guessed the gradient although I am struggling to understand why the x axis is 1/ λ???

Well when you take the gradient it is difference in y over difference in x. Which gives us m= E/(1/ λ) = E λ.
By rearranging the original formula we see E λ = hc. So we know hc =m. I hope that's helpful.
Reply 4
Avatar for Alexokay
Alexokay
OP
11
Original post by Knightwing98124
Well when you take the gradient it is difference in y over difference in x. Which gives us m= E/(1/ λ) = E λ.
By rearranging the original formula we see E λ = hc. So we know hc =m. I hope that's helpful.

Thanks but I think you misread (or I'm being stupid), I understand the gradient part, just dont understand how you got the 1/λ on the x axis as opposed to λ.
It's just to do with the rearranging, you couldn't rearrange it to be identitical to the original equation. E=hc/*λ, if you found the gradient of E against *λ you'd have E/*λ. But it would rearrange with λ on top.
Reply 6
Avatar for Alexokay
Alexokay
OP
11
cheers for helping, but I am failing to understand how it rearranges to for λ to be on top.
E=hc/λ
E/λ = hc (which is the gradient)
If above is correct, I am really struggling to see where the 1/ λ comes from still.
No, if you use 1/*λ as x-axis then the gradient will be E*λ because E÷(1/*λ) = E × *λ = E*λ. So the rearranging will be fine.
Reply 8
Avatar for Sinnoh
Sinnoh
22
Original post by Alexokay
cheers for helping, but I am failing to understand how it rearranges to for λ to be on top.
E=hc/λ
E/λ = hc (which is the gradient)
If above is correct, I am really struggling to see where the 1/ λ comes from still.


E=hcλ≡hc(1λ) E = \dfrac{hc}{\lambda} \equiv hc \left(\dfrac{1}{\lambda}\right)

hc is a constant, and 1λ\frac{1}{\lambda} is your variable

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you