What is the wavelength emitted by compound BB that have spectral line detected at position 12.15 cmcm

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?

λ ≈ 525 nm

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

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