Question

A hydrogen atom is in 4p state.To what state or states can it go by radiating a photon in an allowed transitions?

Answer 1

Answer:

Explanation: energy of n state of hydrogen atom = En = -13.6/n^2eV

energy radiated the transition E = 13.6(1/n^2 - 1/n^2)ev

4th state excite n2 = 5

2nd state excite n1 = 3

E = 13.6(1/3^2 - 1/5^2)

0.967ev

Answer 2

Explanation:

Consider the electron of a hydrogen atom. If the initial state of the electron is given by the quantum numbers $n$ and $l$, the possible final state of the electron is given by the selection rule.

• An electron can transition to any other energy state n'by absorbing or emitting an electron (if $n' > 0$).

• Electrons can change to other energy states $l'=l\pm1$ depending on whether photons are absorbed or emitted. This condition is due to the conservation of angular momentum.

We are given the initial state of the hydrogen atom in 4p sate:

$\implies n_i=4,l_i=1.$

When an atom emits a photon, its angular momentum decreases by $\hbar$ and its orbital angular momentum decreases $l_f=0$. Meanwhile, the electron can end up in any energy level provided that $n > 0.$ Therefore, the final states possible are:

• $1_s \implies n_f=1,l_f=0,$

• $2_s \implies n_f=2,l_f=0,$

• $1_s \implies n_f=3,l_f=0,$

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