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Chapter 27: Problem 42

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

Expert verified
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

01

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 \(f\): \(f = \frac{c}{\lambda} = \frac{3 \times 10^8 \ \mathrm{m/s}}{587.6 \times 10^{-9} \ \mathrm{m}}\)
02

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: \(E = (6.626 \times 10^{-34} \ \mathrm{J \cdot s}) \times (\frac{3 \times 10^8 \ \mathrm{m/s}}{587.6 \times 10^{-9} \ \mathrm{m}})\)
03

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.

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Most popular questions from this chapter

A \(100-\) W lightbulb radiates visible light at a rate of about $10 \mathrm{W} ;$ the rest of the EM radiation is mostly infrared. Assume that the lightbulb radiates uniformly in all directions. Under ideal conditions, the eye can see the lightbulb if at least 20 visible photons per second enter a dark-adapted eye with a pupil diameter of \(7 \mathrm{mm}.\) (a) Estimate how far from the source the lightbulb can be seen under these rather extreme conditions. Assume an average wavelength of 600 nm. (b) Why do we not normally see lightbulbs at anywhere near this distance?
Photons with a wavelength of 400 nm are incident on an unknown metal, and electrons are ejected from the metal. However, when photons with a wavelength of \(700 \mathrm{nm}\) are incident on the metal, no electrons are ejected. (a) Could this metal be cesium with a work function of \(1.8 \mathrm{eV} ?\) (b) Could this metal be tungsten with a work function of 4.6 eV? (c) Calculate the maximum kinetic energy of the ejected electrons for each possible metal when 200 -nm photons are incident on it.
A clean iron surface is illuminated by ultraviolet light. No photoelectrons are ejected until the wavelength of the incident UV light falls below 288 nm. (a) What is the work function (in electron-volts) of the metal? (b) What is the maximum kinetic energy for electrons ejected by incident light of wavelength \(140 \mathrm{nm} ?\)
By directly substituting the values of the fundamental constants, show that the ground state energy for hydrogen in the Bohr model $E_{1}=-m_{\mathrm{e}} k^{2} e^{4} /\left(2 \hbar^{2}\right)$ has the numerical value -13.6 eV.
Photons of wavelength \(350 \mathrm{nm}\) are incident on a metal plate in a photocell and electrons are ejected. A stopping potential of \(1.10 \mathrm{V}\) is able to just prevent any of the ejected electrons from reaching the opposite electrode. What is the maximum wavelength of photons that will eject electrons from this metal?
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