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 501 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?

Question

Jaydan Levy

Physics

17 January, 03:36

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 501 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?

+4

Answers (1)

Know the Answer?

Denise Norton · Answer 1

a) λ = 189.43 10⁻⁹ m b) λ = 269.19 10⁻⁹ m

Explanation:

The diffraction network is described by the expression

d sin θ = m λ

Where m corresponds to the diffraction order

Let's use trigonometry to find the breast

tan θ = y / L

The diffraction spectrum is measured at very small angles, therefore

tan θ = sin θ / cos θ = sin θ

We replace

d y / L = m λ

Let's place in the first order m = 1

Let's look for the separation of the lines (d)

d = λ L / y

d = 501 10⁻⁹ 9.95 10⁻² / 15 10⁻²

d = 332.33 10⁻⁹ m

Now we can look for the wavelength of the other line

λ = d y / L

λ = 332.33 10⁻⁹ 8.55 10⁻²/15 10⁻²

λ = 189.43 10⁻⁹ m

Part B

The compound wavelength B

λ = 332.33 10⁻⁹ 12.15 10⁻² / 15 10⁻²

λ = 269.19 10⁻⁹ m