Question

One line in the helium spectrum is bright yellow and has the wavelength $587.6 \mathrm{nm} .$ What is the difference in energy (in electron-volts) between two helium levels that produce this line?

Short answer

Answer: To find the energy difference, follow these steps: 1. Convert the wavelength to frequency: \(f = \frac{3 \times 10^8 \ \mathrm{m/s}}{587.6 \times 10^{-9} \ \mathrm{m}}\) 2. Calculate the energy difference using Planck's equation: \(E = (6.626 \times 10^{-34} \ \mathrm{J \cdot s}) \times (\frac{3 \times 10^8 \ \mathrm{m/s}}{587.6 \times 10^{-9} \ \mathrm{m}})\) 3. Convert the energy difference to electron-volts: \(E_{\mathrm{eV}} = (6.626 \times 10^{-34} \ \mathrm{J \cdot s}) \times (\frac{3 \times 10^8 \ \mathrm{m/s}}{587.6 \times 10^{-9} \ \mathrm{m}}) \times \frac{6.242 \times 10^{18} \ \mathrm{eV}}{\mathrm{1 \ J}}\) Calculate the final expression to get the energy difference between the two helium levels in electron-volts.

Step-by-step solution

Convert the wavelength to frequency

To convert the wavelength (in nm) to frequency (in Hz), we can use the speed of light equation: \(c = \lambda f\) where \(c\) is the speed of light (\(3 \times 10^8 \ \mathrm{m/s}\)), \(\lambda\) is the wavelength, and \(f\) is the frequency. First, we'll convert the given wavelength from nm to meters: \(\lambda = 587.6 \ \mathrm{nm} \times \frac{1 \ \mathrm{m}}{10^9 \ \mathrm{nm}} = 587.6 \times 10^{-9} \ \mathrm{m}\) Now, we can find the frequency \(f\): \(f = \frac{c}{\lambda} = \frac{3 \times 10^8 \ \mathrm{m/s}}{587.6 \times 10^{-9} \ \mathrm{m}}\)

Calculate the energy difference

We can find the energy difference between the two levels using Planck's equation: \(E = hf\) where \(E\) is the energy difference, \(h\) is the Planck's constant (\(6.626 \times 10^{-34} \ \mathrm{J \cdot s}\)), and \(f\) is the frequency. Substitute the values of \(h\) and \(f\) into the equation: \(E = (6.626 \times 10^{-34} \ \mathrm{J \cdot s}) \times (\frac{3 \times 10^8 \ \mathrm{m/s}}{587.6 \times 10^{-9} \ \mathrm{m}})\)

Convert the energy difference to electron-volts

Finally, we need to convert the energy difference from joules to electron-volts (eV) using the conversion factor: \(1 \ \mathrm{J} = 6.242 \times 10^{18} \ \mathrm{eV}\) \(E_{\mathrm{eV}} = E \times \frac{6.242 \times 10^{18} \ \mathrm{eV}}{\mathrm{1 \ J}}\) Now, we can calculate the energy difference in electron-volts: \(E_{\mathrm{eV}} = (6.626 \times 10^{-34} \ \mathrm{J \cdot s}) \times (\frac{3 \times 10^8 \ \mathrm{m/s}}{587.6 \times 10^{-9} \ \mathrm{m}}) \times \frac{6.242 \times 10^{18} \ \mathrm{eV}}{\mathrm{1 \ J}}\) After calculating the above expression, we get the energy difference between two helium levels that produce the bright yellow line.