Passing an electric current through an unknown gas produces several distinct wavelengths of visible light. In this problem, you will find the wavelengths of the unknown spectrum by observing that they form first-order maxima at angles of 24.2°, 30.5°, 35.3°, and 41.7° when projected on a diffraction grating having 10,000 lines per centimeter.
(a) What is the wavelength (in nm) for the line at 23.80°?
(b) What is the wavelength (in nm) for the line at 30.20°?

1)At 24.2°, 30.5°, 35.3°, and 41.7°, the wavelengths are 409.9 nm, 507.5 nm, 577.9 nm and 665.2 nm respectively.

A) at 23.80°, wavelength = 403.5 nm

B) at 30.2°,wavelength = 503 nm

Explanation:

1) Order of maxima in the question is m=1

θ1 = 24.2°

θ2 = 30.5°

θ3 = 35.3°

θ4 = 41.7°

Equation for mth order maxima is;

d sinθ = mλ

Where d is the separation lines between grating and λ is the wavelength

From the question, the grating has 10,000 lines per cm

Thus, d = 1/10000 = 0.0001 cm then converting it to meters, d= 0.0001 x 10^-(2) = 10^-(6)

So,

Wavelength(λ) for θ1 = 24.2°;

dsinθ = mλ

m = 1 and so :λ = dsinθ

λ = 10^-(6) x sin 24.2 = 409.9 x 10^(-9) m or simply 409.9 nm

Wavelength(λ) for θ2 = 30.5°;

dsinθ = λ;

λ = 10^-(6) x sin 30.5 = 507.5 nm

Wavelength(λ) for θ3 = 35.3°;

dsinθ = λ;

λ = 10^-(6) x sin 35.3 = 577.9 nm

Wavelength(λ) for θ4 = 41.7°;

dsinθ = λ;

λ = 10^-(6) x sin 41.7 = 665.2 nm

A) At 23.8°;

dsinθ = λ;

λ = 10^-(6) x sin sin 23.8 = 403.5 nm

B) At 30.2°;

dsinθ = λ;

λ = 10^-(6) x sin sin 23.8 = 503 nm

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explanation:

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