Complete question is;
A chemist identifies compounds by identifying bright lines in their spectra. She does so by heating the compounds until they glow, sending the light through a diffraction grating, and measuring the positions of first-order spectral lines on a detector 15.0 cm behind the grating. Unfortunately, she has lost the card that gives the specifications of the grating. Fortunately, she has a known compound that she can use to calibrate the grating. She heats the known compound, which emits light at a wavelength of 461 nm, and observes a spectral line 9.95 cm from the center of the diffraction pattern.
What is the wavelength emitted by compound B that have spectral line detected at position and 12.15 cm?
Answer:
λ ≈ 525 nm
Step-by-step explanation:
The diffraction grating equation of n-th order is given by;
nλ = dsinθₙ
For 1st order, n=1. Thus;L, we have;
λ = dsinθ₁
We are told that the first-order spectral lines on a detector 15.0 cm behind the grating. Also that, she observes a spectral line 9.95 cm from the center of the diffraction pattern.
Therefore, the angle of the known compound's 1st order is given by the relation:
tanθ₁ = 9.95/15
tanθ₁ = 0.6633
θ₁ = tan⁻¹(0.6633)
θ₁ = 33.56º
We are told that the unknown compound emits light at a wavelength of 461 nm.
Thus, from the first order equation above, we have;
d = λ/sinθ₁
d = 461nm/sin(33.56º)
d = 461nm/0.55281
d ≈ 834nm
λ = dsinθ₁= 870nm sin(29.68º) = 431nm
Now, spectral line is given as 12.15 cm from the center of the diffraction pattern. Thus;
θ₁ = tan⁻¹(12.15/15)
θ₁ = tan⁻¹(0.81)
θ₁ = 39°
Since λ = dsinθ₁
Thus;
λ = 834sin(39°)
λ = 834 × 0.6293
λ ≈ 525 nm