a diffration grating has 250lines/cm. the 4th order image is formed at an angle of 4.00. what is the wavelength ?
the formula to use is wavelenth=dsin/n , but i am doing somthing wrong i thik
do we do simthing to 250?
We have 250 lines per cm. How many cm per line?
what u mean
d is the distance between lines. If you have 250 lines per centimeter, what is the distance between the lines?
1/250?
Yep. Try using that for d, and you should be good to go!
i did that but still
is n 4 ?
If they are using the standard notation, yeah. The center fringe is the 0th order, next out is 1st, etc.
the a answer is 698nm
what u did was 1/250*sin4/4
Remember, 1/250 is in cm.
so first its 1/250 * 1000?
what do we change it to ?
1/250 * 10^(-2) m
i still dont get 698nm
Strange. I do. Make sure your calculator is working with degrees (not radians) \[\frac{1}{250} \times \sin(4^o) \times \frac{1}{4}\]
You should end up with 6.97565*10^(-07) m, which is (rounded to) 678 nm
oh okey i got that to before and i thought i was wrong
wait so how does that work ? how do u just round it ?
Ah \[6.97565\times 10^{-7} = 697.565\times 10^{-9}\] \[10^{-9}\textrm{m} = 1\textrm{nm}\]
so would it have been easier if i had converted 250cm to nm first?
just put 10^-9 after 250
1/250 cm to nanometers would have worked, yeah. "Easier" depends on how you look at it. I prefer to take everything to base units for situations like this.
No, not 10^(-9) 1/250 cm = 1/250 * 10^7 nm
You should look up a chart on metric conversions. Keep that around until you're more familiar with using it.
i have a chart
it says -9
10^(-9)m = nm or 10^9 nm = m nm are smaller, so there will be more of them.
can u help me with other questions
Sure
hold on
a photon with a wavelenth of 950nm has what frequecy
We know that \[c = \lambda \nu \] where c is the speed of light, lambda (the kind of h looking thing) is the wavelength, and nu (the v looking thing) is the frequency. We know the speed of light is 3.0*10^8 m/s, so using the frequency, we can find the wavelength (be careful of units!!!)
the next question is the same but this time it says whats the energy ?
For a photon \[E = h \nu\] where h is Planck's constant
i have e=pc on my sheet
wait in the first question it was asked frequency their is no f in the equation u gave me
\[p = \frac{h}{\lambda} = \frac{h\nu}{c}\] So \[E = pc = \frac{h\nu}{c}c = h\nu\]
What you call c, I'm calling nu. I'll switch from here on out. \[f = \nu\]
err, f. Not c.
oh ok
wait u mean no v but f
c is a constant and we need it
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