Question

The wavelength of the blue light emitted from a hydrogen atom is \(410.1 \mathrm{~nm}\). This light is a result of electron transitions from the \(n=5\) to the \(n=2\) energy levels. How much higher in energy is the \(n=5\) energy level than the \(n=2\) energy level?

Short answer

The \(n=5\) energy level is \(4.84 \times 10^{-19}\) J higher than the \(n=2\) energy level.

Step-by-step solution

Conversion of wavelength to metres

Given the wavelength of the blue light is 410.1 nm. Convert nanometres to metres by multiplying by \(1.0 \times 10^{-9}\), so our wavelength in metres is \(410.1 \times 1.0 \times 10^{-9} = 4.101 \times 10^{-7}\) m.

Calculation of frequency

Now, calculate the frequency of the emitted light using the formula for the speed of light in a vacuum \(c = λν\), where \(c\) is the speed of light, \(λ\) is the wavelength and \(ν\) is the frequency. Rearrange the equation and solve for \(ν = c/λ\), therefore \(ν = 3.00 \times 10^8 m/s / 4.101 \times 10^{-7} m = 7.31 \times 10^{14} Hz\).

Calculation of energy difference

Finally, calculate the energy difference by using the formula for the energy of a wave, \(E = hv\), where \(h\) is Planck's constant and \(ν\) is the frequency. Therefore, \(E = 6.63 \times10^{-34} J s \times 7.31 \times 10^{14} Hz = 4.84 \times 10^{-19} J\).

Key concepts

Wavelength Conversion

When working with light wavelengths, it's important to understand how units are converted. Often, wavelengths are presented in nanometers (nm) because these units are convenient for light in the visible spectrum. However, calculations in physics often require us to convert this value to meters (m) for consistency with other units.

To convert nanometers to meters, we use the conversion factor where 1 nm equals \(1 \times 10^{-9}\) meters. Thus, to convert a wavelength from nanometers to meters, multiply the nanometer value by \(1 \times 10^{-9}\).

For example, if we have a wavelength of blue light at 410.1 nm, we calculate:

• \(410.1 \times 1 \times 10^{-9} = 4.101 \times 10^{-7} \ \text{m}\)

This conversion ensures that we can use the wavelength in a variety of scientific equations accurately and consistently.

Frequency Calculation

Calculating the frequency of light involves using a fundamental principle in physics that relates the speed of light to its wavelength and frequency. The formula used is:

\[c = \lambda u\]
where:

• \(c\) is the speed of light in a vacuum, approximately \(3.00 \times 10^8\ \text{m/s}\)
• \(\lambda\) is the wavelength in meters
• \(u\) is the frequency in hertz (Hz)

To find the frequency (\(u\)), rearrange the formula:
\[u = \frac{c}{\lambda}\]

Using our earlier calculation of a wavelength of \(4.101 \times 10^{-7}\) meters for blue light, the frequency calculation proceeds as:

• \(u = \frac{3.00 \times 10^8\ \text{m/s}}{4.101 \times 10^{-7}\ \text{m}} = 7.31 \times 10^{14}\ \text{Hz}\)

This result shows how often the wave cycles pass a given point per second.

Energy Difference Calculation

The energy of a photon can be determined using its frequency through Planck’s equation:

\[E = hu\]
Where:

• \(E\) is the energy in joules (J)
• \(h\) is Planck’s constant, approximately \(6.63 \times 10^{-34}\ \text{J s}\)
• \(u\) is the frequency in hertz (Hz)

Utilizing the frequency we calculated, \(7.31 \times 10^{14}\) Hz, the energy can be calculated as:

• \(E = 6.63 \times 10^{-34}\ \text{J s} \times 7.31 \times 10^{14}\ \text{Hz} = 4.84 \times 10^{-19}\ \text{J}\)

This energy difference corresponds to the amount of energy needed for the electron to transition between the specified energy levels (from \(n=5\) to \(n=2\)) in a hydrogen atom. Understanding and calculating these transitions help us comprehend fundamental atomic interactions.