Hello everyone! I was wondering if anyone can help me with this question? Picture attached.

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

Hello everyone! I was wondering if anyone can help me with this question? Picture attached.

Chemistry
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

\[\lambda=\frac{hc}{E_{ph}}\] right?
Do I get the difference of eV, e.g. like when n=2 goes to n=1, eV=-13.6-(-3.4)?

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

the lowest the electron fall the higher is the energy that release, the higher the energy the lower the wavelength. If an electron fall from any level to the level 1 will release the most energy (lower wavelength), then if it felt from the 6 to the 1 release more energy than if it fell from the 5. And if it felt from the 5 to the 1 release more than if does from the 4 to the 1. Then if an electron fall to the n=2 wherever level is coming from will always release less energy that any other electron falling to the n=1
4 ->3 4 ->2 2 ->1 3 -> 1
Thank you so much! I understand it now.
Yes you can replace the values in the formula and calculate the wavelenght
Thank you! @Cuanchi
https://en.wikipedia.org/wiki/Hydrogen_spectral_series
Welcome To OpenStudy! Here you will find great helpers and friends, a community of students that help students! We hope you enjoy the experience!

Not the answer you are looking for?

Search for more explanations.

Ask your own question