how does the value of n of an orbit in the Bohr model of hydrogen relate to the

Question

how does the value of n of an orbit in the Bohr model of hydrogen relate to the energy of an electron in that orbit?

In the visible portion of the atomic emission spectrum of hydrogen, are there any bright lines due to electron transitions to the groud state?

In what ways should the emission spectra of H and He+ be alike and in what ways should be different?

Why do matter waves not add significantly to the challenge of hitting a baseball thrown at 99mph (44m/s)?

How does de Broglie's hypothesis that electrons behave like waves explain the stability of the electron orbits in a hydrogen atom?

Explanation / Answer

1) E = -13.6/n2 eV

2) Ground state corresponds to n=1. Transition to this state emits UV radiation. Transition to n=2 emits visible light.

3) He+ and H both has only one electron orbitting about the atomic nucleus. Therefore the wavelength of emitted radiation is due only to one electron transition.

The difference is in the Energy or wave length of the emitted radiation. as E is directly proportional and wave length is inversely proportional to square of atomic number.

4) Because the uncertainity in position measurement is too little to notice, because of size of the base ball.

5) If electrons are particles, they would be attracted by the positively charged central nucleus. Instead if it is a wave then it can spread around the nucleus like a cloud with intensity decreasing as it moves away from the centre.

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