Question

From the data in Table below, what photon wavelength can ionize Oxygen? What wavelength can ionize Silicon? TABLE 24.1 Composition of a Typical Star, Our Sun Element Number of Protons (Atomic Number) Relative Number of Atoms Percentage by Mass Energy to Remove Outermost Electron Hydrogen 1 10^12 71.1% 2.179 x 10^-18 J Helium 2 9.64 x 10^10 27.4% 3.940 x 10^-18 J Carbon 6 2.88 x 10^8 0.25% 1.804 x 10^-18 J Nitrogen 7 7.94 x 10^7 0.08% 2.328 x 10^-18 J Oxygen 8 5.75 x 10^8 0.65% 2.182 x 10^-18 J Neon 10 8.91 x 10^7 0.13% 3.454 x 10^-18 J Silicon 14 4.07 x 10^7 0.06% 1.306 x 10^-18 J Iron 26 3.47 x 10^7 0.14% 1.266 x 10^-18 J Gold 79 8 0.00000011% 1.478 x 10^-18 J Uranium 92 0.4 <0.000000007% 0.992 x 10^-18 J The wavelength of a photon that can ionize Oxygen is ?Oxygen = [ ] nm. The wavelength of a photon that can ionize Silicon is ?Silicon = [ ] nm. 

          From the data in Table below, what photon wavelength can ionize Oxygen? What wavelength can ionize Silicon?

TABLE 24.1 Composition of a Typical Star, Our Sun
Element Number of Protons (Atomic Number) Relative Number of Atoms Percentage by Mass Energy to Remove Outermost Electron
Hydrogen 1 10^12 71.1% 2.179 x 10^-18 J
Helium 2 9.64 x 10^10 27.4% 3.940 x 10^-18 J
Carbon 6 2.88 x 10^8 0.25% 1.804 x 10^-18 J
Nitrogen 7 7.94 x 10^7 0.08% 2.328 x 10^-18 J
Oxygen 8 5.75 x 10^8 0.65% 2.182 x 10^-18 J
Neon 10 8.91 x 10^7 0.13% 3.454 x 10^-18 J
Silicon 14 4.07 x 10^7 0.06% 1.306 x 10^-18 J
Iron 26 3.47 x 10^7 0.14% 1.266 x 10^-18 J
Gold 79 8 0.00000011% 1.478 x 10^-18 J
Uranium 92 0.4 <0.000000007% 0.992 x 10^-18 J

The wavelength of a photon that can ionize Oxygen is ?Oxygen = [ ] nm.
The wavelength of a photon that can ionize Silicon is ?Silicon = [ ] nm.
Show more…
From the data in Table below, what photon wavelength can ionize Oxygen? What wavelength can ionize Silicon? TABLE 24.1 Composition of a Typical Star, Our Sun Element Number of Protons (Atomic Number) Relative Number of Atoms Percentage by Mass Energy to Remove Outermost Electron Hydrogen 1 10^12 71.1% 2.179 x 10^-18 J Helium 2 9.64 x 10^10 27.4% 3.940 x 10^-18 J Carbon 6 2.88 x 10^8 0.25% 1.804 x 10^-18 J Nitrogen 7 7.94 x 10^7 0.08% 2.328 x 10^-18 J Oxygen 8 5.75 x 10^8 0.65% 2.182 x 10^-18 J Neon 10 8.91 x 10^7 0.13% 3.454 x 10^-18 J Silicon 14 4.07 x 10^7 0.06% 1.306 x 10^-18 J Iron 26 3.47 x 10^7 0.14% 1.266 x 10^-18 J Gold 79 8 0.00000011% 1.478 x 10^-18 J Uranium 92 0.4 <0.000000007% 0.992 x 10^-18 J The wavelength of a photon that can ionize Oxygen is ?Oxygen = [ ] nm. The wavelength of a photon that can ionize Silicon is ?Silicon = [ ] nm.

Added by Shannon S.

Close

University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
AceChat toggle button
Close icon
Ace pointing down

Please give Ace some feedback

Your feedback will help us improve your experience

Thumb up icon Thumb down icon
Thanks for your feedback!
Profile picture
From the data in Table 24.1, what photon wavelength can ionize Oxygen? What wavelength can ionize Silicon? TABLE 24.1 Composition of a Typical Star, Our Sun Element Number of Protons (Atomic Number) Relative Number of Atoms Percentage by Mass Energy to Remove Outermost Electron Hydrogen 1 10^12 71.1% 2.179 x 10^-18 J Helium 2 9.64 x 10^10 27.4% 3.940 x 10^-18 J Carbon 6 2.88 x 10^8 0.25% 1.804 x 10^-18 J Nitrogen 7 7.94 x 10^7 0.08% 2.328 x 10^-18 J Oxygen 8 5.75 x 10^8 0.65% 2.182 x 10^-18 J Neon 10 8.91 x 10^7 0.13% 3.454 x 10^-18 J Silicon 14 4.07 x 10^7 0.06% 1.306 x 10^-18 J Iron 26 3.47 x 10^7 0.14% 1.266 x 10^-18 J Gold 79 8 0.00000011% 1.478 x 10^-18 J Uranium 92 0.4 <0.000000007% 0.992 x 10^-18 J The wavelength of a photon that can ionize Oxygen is λOxygen = nm. The wavelength of a photon that can ionize Silicon is λSilicon = nm.
Close icon
Play audio
Feedback
Powered by NumerAI
David Collins Ivan Kochetkov
Danielle Fairburn verified

Dominador Tan and 72 other subject Physics 101 Mechanics educators are ready to help you.

Ask a new question
*

Labs

-

Want to see this concept in action?

NEW

Explore this concept interactively to see how it behaves as you change inputs.

View Labs
*

Key Concepts

-
Key Concept
Premium Feature
Explore the core concept behind this problem.
Play button
Key Concept
Premium Feature
Explore the core concept behind this problem.
Your browser does not support the video tag.
*

Recommended Videos

-
table-1-hydrogen-gas-emission-spectrum-wavelength-410-nm-violet-spectral-iing-blue-violet-spectral-line-435-nm-blue-green-spectrall-ilne-485-nmi-red-spectral-iine-660-nm-the-change-in-cnerey-27534

Table 1: Hydrogen gas emission spectrum Wavelength: 410 nm Violet spectral line Blue-violet spectral line 435 nm Blue-green spectral line 485 nm Red spectral line: 660 nm The change in energy of the atom is equal in magnitude to the energy of the photon emitted, ΔEatom = ΔEphoton. So, first, use the equation ΔE = hc/Ι to find the energy of the photon and therefore the change in energy of the atom: Planck's constant, h, is equal to 6.626*10^-34 and the speed of light is 3.00*10^8 m/s. Make sure you use the wavelength you identified above for the blue-green spectral line in the hydrogen emission spectrum and convert the wavelength from nanometers to meters. ΔEatom should be a positive number. The magnitude of ΔEatom for the blue-violet wavelength identified above? The magnitude of ΔEatom for the blue-green wavelength identified in part 2? The magnitude of ΔEatom for the red wavelength identified in part 3?

Snehal J.
table-view-list-view-helium-gas-emission-spectrum-theoretical-wavelength-nm-energy-j-44610-19j-447-color-violet-398710-19j0-471-blue-396710-19j0-501-green-319910-19j0-281110-191j-587-yellow-80804

Hel iu m g a s e m i s s i o n s p e c t r u m T h e o r e t i c a l e n e r g y ( J ) W a v e l e n g t h ( n m ) C o l o r 4 . 4 4 6 * 1 0 ^ ( - 1 9 ) J 4 4 7 v i o l e t 3 . 9 8 7 * 1 0 ^ ( - 1 9 ) J 4 7 1 b l u e 3 . 9 6 7 * 1 0 ^ ( - 1 9 ) J 5 0 1 g r e e n 3 . 1 9 9 * 1 0 ^ ( - 1 9 ) J 5 8 7 y e l l o w 2 . 8 1 1 * 1 0 ^ ( - 1 9 ) J 6 6 8 r e d T h e e n e r g y o f e a c h o b s e r v e d w a v e l e n g t h m a y b e c a l c u l a t e d u s i n g t h e f o l l o w i n g r e l a t i o n s h i p E = h c / λ . P l a n k ' s c o n s t a n t , h , i s e q u a l t o 6 . 6 2 6 ! 1 0 ^ - 3 4 J ! s a n d t h e s p e e d o f l i g h t i s 3 . 0 0 ! 1 0 ^ 8 m / s . M a k e s u r e y o u c o n v e r t t h e w a v e l e n g t h f r o m n a n o m e t e r s t o m e t e r s .

Ma Ednelyn L.
the-emission-spectrum-of-hydrogen-has-bright-blue-green-line-at-486-nn-this-electron-ends-in-the-energy-level-what-the-initial-energy-of-the-ectron-that-produced-this-photon-find-the-energy-70837

The emission spectrum of Hydrogen has a bright blue-green line at 486 nm. This electron ends in the n_f = 2 energy level. What is the initial energy of the electron that produced this photon? (Find the energy of the electron in its excited state, n_i). An electron in the n = 2 state of hydrogen is excited by absorbing the longest possible wavelength photon. What is the wavelength of this photon and what is its energy?

Shyam P.
*

Recommended Textbooks

-
University Physics with Modern Physics

University Physics with Modern Physics

Hugh D. Young 14th Edition
achievement 1,580 solutions
Physics: Principles with Applications

Physics: Principles with Applications

Douglas C. Giancoli 7th Edition
achievement 1,228 solutions
Fundamentals of Physics

Fundamentals of Physics

David Halliday, Robert Resnick , Jearl Walker 10th Edition
achievement 1,101 solutions
*

Transcript

-
00:01 For oxygen, energy to remove outermost electrons, we have e is equal to 2 .182 times 10, raised to the part of negative 18 joules.
00:12 This is 18 joules.
00:15 So you're going to use e that is equal to h times c if i divide the lambda.
00:25 So the lambda is equals to h times c, divided by e photon.
00:34 So we have equal to 6 .626 times 10 raised to the power of negative 34 times 3 times 10 raised to the power of 8.
00:45 This is positive 8.
00:50 Divide divide 2 .182 times 10 raised to the part of negative 18...
Need help? Use Ace
Ace is your personal tutor. It breaks down any question with clear steps so you can learn.
Start Using Ace
Ace is your personal tutor for learning
Step-by-step explanations
Instant summaries
Summarize YouTube videos
Understand textbook images or PDFs
Study tools like quizzes and flashcards
Listen to your notes as a podcast
Continue solving this problem
Create a free account to:
  • View full step-by-step solution
  • Ask follow-up questions with Ace AI
  • Save progress and study later
Continue Free
Numerade

Get step-by-step video solution
from top educators

Continue with Clever
or



By creating an account, you agree to the Terms of Service and Privacy Policy
Already have an account? Log In

A free answer
just for you

Watch the video solution with this free unlock.

Numerade

Log in to watch this video
...and 100,000,000 more!


EMAIL

PASSWORD

OR
Continue with Clever