College

The experimental setup used to determine the wavelengths of light emitted by a hydrogen atom is shown below. A student uses a grating with 29,426 lines per inch and sets up the grating so that it is 125.0 cm from the lamp. They measure a spectral line to be at 99.9 cm from the lamp. Recall Bragg’s equation, where [tex] y = d \sin \theta [/tex].

a) What is the wavelength of light (in nanometers) for this spectral line?

b) Is this light in the visible region of the electromagnetic spectrum?

Answer :

(a) The wavelength of light is approximately 540 nm.

(b) This wavelength is in the visible range of the electromagnetic spectrum.

To determine the wavelength of the light emitted by the hydrogen atom in this experimental setup, we can use Bragg's equation, which states:

[tex]y = d \sin \theta[/tex]

Where:

[tex]y[/tex] is the distance from the grating to the spectral line,

[tex]d[/tex] is the distance between the grating lines,

[tex]\theta[/tex] is the angle of diffraction.

Step 1: Calculate the Distance Between Grating Lines (d) -

The grating has 29,426 lines per inch. To find the grating spacing in centimeters, we first convert lines per inch to lines per centimeter:

[tex]29,426 \text{ lines/inch} \times \frac{1 \text{ inch}}{2.54 \text{ cm}} \approx 11,566 \text{ lines/cm}[/tex]

Next, we calculate the spacing (d):

[tex]d = \frac{1 \text{ cm}}{11,566 \text{ lines/cm}} \approx 8.64 \times 10^{-5} \text{ cm}[/tex]
(We can convert this to meters for convenience later if needed.)

Step 2: Set Up the Geometry for Our Measurements -

The distance from the lamp to the grating is given as 125.0 cm, and the measurement to the spectral line is 99.9 cm. The angle [tex]\theta[/tex] can be calculated using:

[tex]\tan \theta = \frac{y}{L}[/tex]

Where L is the distance from the lamp to the grating, and y is the distance to the spectral line (99.9 cm).

[tex]\theta = \tan^{-1} \left( \frac{99.9 \text{ cm}}{125.0 \text{ cm}} \right) \approx \tan^{-1} (0.7992) \approx 38.7^\circ[/tex]

Step 3: Calculate the Wavelength [tex]\lambda[/tex] -

Now apply Bragg's equation. We substitute the values for d and sin([tex]\theta[/tex]):

[tex]\lambda = d \sin \theta[/tex]

Calculating [tex]\sin(38.7^\circ)[/tex]:

[tex]\sin(38.7^\circ) \approx 0.6249[/tex]

Then substituting these into the equation:

[tex]\lambda = 8.64 \times 10^{-5} \text{ cm} \times 0.6249 \approx 5.398 \times 10^{-5} \text{ cm}[/tex]

Convert this to nanometers (1 cm = 10⁷ nm):

[tex]\lambda \approx 5.398 \times 10^{-5} \text{ cm} \times 10^7 \text{ nm/cm} = 539.8 \text{ nm}[/tex]

Final Result for Part (a) -

The wavelength of the spectral line measured is approximately 540 nm.

Part (b): Is this light in the visible region?

Yes, the wavelength of 540 nm falls within the visible spectrum, which ranges approximately from 380 nm to 750 nm. Therefore, this light is visible to the human eye.

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