What Angle Will a Hydrogen Spectral Line Appear in 2nd Order?

[smaan]

Homework Statement

Hello,

I have a question regarding hydrogen spectral emission.

A hydrogen source is viewed with a grating spectrometer, one spectral lines occurs at 20.5 degrees in the 1st order. What angle will this line appear in 2nd order (viewed through same spectrometer)

a) cannot find this angle without ruling spacing of grating
b) cannot find this angle without wavelength of spectral line
c) it IS possible to find this angle with the given info



Please help!

Homework Equations

d(sinθ)=mλ

The Attempt at a Solution

When I think about this, do we not need the ruling spacing to find the wavelength? From there, can we not find the 2nd order angle? The problem is, there is no option for A and B. And that's assuming I'm on the right track.

 

[ehild]

There is the equation d(sinθ)=mλ, and given that θ=20.5 degrees when m=1. Can you find λ/d?
The question is the angle when m=2. Do you need anything else but λ/d?


ehild

 

[smaan]

There is the equation d(sinθ)=mλ, and given that θ=20.5 degrees when m=1. Can you find λ/d?
The question is the angle when m=2. Do you need anything else but λ/d?


ehild

d is the space between the grating, correct? And λ is the wavelength. I don't have either. Just an angle for m=1. Still confused, but thank you for the reply.

 

[BvU]

Why this confusion? You don't have either, but you do have the ratio ! And what do you really need ? Aha!

 

[smaan]

So in my case the angle appearing at the 2nd order will be 41.0 degrees? Is that correct?

 

[BvU]

No. Just try to imagine where the fifth order would end up!

No that's corny. You have an equation for this $\theta$. Solve it.

 

[smaan]

Yes, it would be over 90°, makes no sense. Which leads me to believe I need grating. So would my answer be

a) cannot find this angle without ruling spacing of grating

?

 

[BvU]

No. It's just that if $\sin \theta = q$, that doesn't mean that $\sin 2\theta = 2q$.
You have q from the 20.5 $^\circ$, now find $\theta_2$!

 

[smaan]

sinθ = q
sin20.5 = q
0.350207 = q

0.350207*2 = 0.700415

sin^-1(0.700415) = 44.5°

Is that correct?

By the way, thank you for all you're help, BvU. I appreciate you're time.

 

[BvU]

I had the same answer... So either it's right or we are both wrong.

 

[BvU]

Thanks for all the thanks, but, really, once is more than adequate...