Chemistry, asked by shoeb92, 10 months ago

In excited hydrogen atom electron return from 7 energy level to ground state in 4 hydrogen atom, find the maximum number of photons that can be emitted during dexcitening ?​

Answers

Answered by Anonymous
18

Answer:

Since comments caused certain level of confusion, I guess I'll try to provide a further illustration. You should consider all possibilities for an electron "jumping" down the excited energy state nn to the ground state n=1n=1. Electron doesn't get stuck forever on any of the levels with n>1n>1.

Besides that, spectra is not a characteristic of a single excited atom, but an ensemble of many and many excited hydrogen atoms. In some atoms electrons jump directly from n=6n=6 to n=1n=1, whereas in some others electrons undergo a cascade of quantized steps of energy loss, say, 6→5→16→5→1 or 6→4→2→16→4→2→1. The goal is to achieve the low energy state, but there is a finite number of ways NN of doing this.

Answered by Anonymous
6

Answer:

The number of spectral lines produced when an electron in the nthlevel drops down to the ground state is given by n(n-1)/2. The energy associated with the first orbit in the hydrogen atom is –2.18×10-18 J atom-1.

When the excited electron of an H atom in n =6 drops to the ground state, the following transitions are possible: Hence, a total numberof (5 + 4 + 3 + 2 + 1) 15 lines will be obtained in the emission spectrum.

Hope this help u ☺!!

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