Achemist 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.part a: what is the wavelength emitted by compound a that have spectral line detected at position 8.55 cm? part b: what is the wavelength emitted by compound b that have spectral line detected at position and 12.15 cm?
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Ответ:
PART A: 412.98 nm
PART B: 524.92 nm
Explanation:
The equation below can be used for a diffraction grating of nth order image:
n*λ = d*sinθ
Therefore, for first order images, n = 1 and:
λ = d*sinθ.
The angle θ can be calculated as follow:
tan θ = 9.95 cm/15.0 cm = 0.663 and
θ = (0.663) = 33.56°
Thus: d =λ/sin θ = 461/sin 33.56° = 833.97 nm
PART A:
For a position of 8.55 cm:
tan θ = 8.55 cm/15.0 cm = 0.57 and
θ = (0.57) = 29.68°
Therefore:
λ =d*sin θ = 833.97*sin 29.68° = 412.98 nm
PART B:
For a position of 12.15 cm:
tan θ = 12.15 cm/15.0 cm = 0.81 and
θ = (0.81) = 39.01°
Therefore:
λ =d*sin θ = 833.97*sin 39.01° = 524.92 nm
Ответ:
we need a picture
Explanation:
but you can ask the question again with one