Question

analyzing data
interpret emission spectra

each color of light has a unique wavelength and frequency. some sources of light produce only specific wavelengths. for example, when you pass an electrical current through a sample of hydrogen gas, the hydrogen emits a mixture of light that appears pink. every element has its own \signature\ of emitted light wavelengths, called an emission spectrum. an element’s emission spectrum may contain a few or many wavelengths of visible light.

other sources, including light bulbs and the sun, emit all colors of light, which is observed as white light. you can demonstrate this phenomenon by using a glass prism to \split\ sunlight into a continuous spectrum of all the colors of visible light. this is the same effect you see in a rainbow.

the table shows some of the wavelengths of light emitted by hydrogen and other elements. the speed of a light wave is related to its frequency and wavelength by this mathematical formula:

wave speed = frequency × wavelength

Explanation:

Assuming the problem is to calculate the speed of light wave for each element using the formula \( \text{wave speed} = \text{frequency} \times \text{wavelength} \), here is the step - by - step solution for one of the elements (taking hydrogen with wavelength \( 4.34\times10^{-7}\space m \) and frequency \( 6.91\times10^{14}\space Hz \) as an example):

Step 1: Identify the formula and values

We use the formula \( v = f\times\lambda \), where \( v \) is the wave speed, \( f \) is the frequency, and \( \lambda \) is the wavelength. Here, \( f = 6.91\times 10^{14}\space Hz \) and \( \lambda=4.34\times 10^{-7}\space m \)

Step 2: Substitute the values into the formula

\( v=(6.91\times 10^{14})\times(4.34\times 10^{-7}) \)
Using the rule of exponents \( a^{m}\times a^{n}=a^{m + n} \) and multiplying the coefficients: \( 6.91\times4.34 = 29.9894 \) and \( 10^{14}\times10^{-7}=10^{14 - 7}=10^{7} \)
So \( v = 29.9894\times 10^{7}=2.99894\times 10^{8}\space m/s \) (approximate value of the speed of light)

If we do this for all the elements:

For hydrogen (violet):

Step 1: Formula and values

\( v = f\times\lambda \), \( f = 6.91\times 10^{14}\space Hz \), \( \lambda = 4.34\times 10^{-7}\space m \)

Step 2: Calculation

\( v=(6.91\times 10^{14})\times(4.34\times 10^{-7})=6.91\times4.34\times10^{14 - 7}=29.9894\times 10^{7}\approx3.0\times 10^{8}\space m/s \)

For hydrogen (blue):

Step 1: Formula and values

\( v = f\times\lambda \), \( f = 6.17\times 10^{14}\space Hz \), \( \lambda = 4.86\times 10^{-7}\space m \)

Step 2: Calculation

\( v=(6.17\times 10^{14})\times(4.86\times 10^{-7})=6.17\times4.86\times10^{14 - 7}=29.9862\times 10^{7}\approx3.0\times 10^{8}\space m/s \)

For helium (green):

Step 1: Formula and values

\( v = f\times\lambda \), \( f = 5.93\times 10^{14}\space Hz \), \( \lambda = 5.06\times 10^{-7}\space m \)

Step 2: Calculation

\( v=(5.93\times 10^{14})\times(5.06\times 10^{-7})=5.93\times5.06\times10^{14 - 7}=29.9958\times 10^{7}\approx3.0\times 10^{8}\space m/s \)

For sodium (yellow):

Step 1: Formula and values

\( v = f\times\lambda \), \( f = 5.27\times 10^{14}\space Hz \), \( \lambda = 5.69\times 10^{-7}\space m \)

Step 2: Calculation

\( v=(5.27\times 10^{14})\times(5.69\times 10^{-7})=5.27\times5.69\times10^{14 - 7}=29.9863\times 10^{7}\approx3.0\times 10^{8}\space m/s \)

For neon (orange):

Step 1: Formula and values

\( v = f\times\lambda \), \( f = 4.98\times 10^{14}\space Hz \), \( \lambda = 6.03\times 10^{-7}\space m \)

Step 2: Calculation

\( v=(4.98\times 10^{14})\times(6.03\times 10^{-7})=4.98\times6.03\times10^{14 - 7}=29.9994\times 10^{7}\approx3.0\times 10^{8}\space m/s \)

For hydrogen (red):

Step 1: Formula and values

\( v = f\times\lambda \), \( f = 4.57\times 10^{14}\space Hz \), \( \lambda = 6.56\times 10^{-7}\space m \)

Step 2: Calculation

\( v=(4.57\times 10^{14})\times(6.56\times 10^{-7})=4.57\times6.56\times10^{14 - 7}=29.9792\times 10^{7}\approx3.0\times 10^{8}\space m/s \)

Answer:

The speed of light wave for each element is approximately \( \boldsymbol{3.0\times 10^{8}\space m/s} \) (which is consistent with the known speed of light in a vacuum, indicating that light waves from these emission spectra travel at the speed of light)