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(B) The energy of an electron in the ground state $(n = 1)$ of a hydrogen atom is $E_1 = -13.6 \, eV$.When energy $E_{given} = 12.4 \, eV$ is supplied

Vedclass · Accepted Answer

$n = 3$

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(B) The energy of an electron in the ground state $(n = 1)$ of a hydrogen atom is $E_1 = -13.6 \, eV$.When energy $E_{given} = 12.4 \, eV$ is supplied,the new energy $E_n$ of the electron will be:$E_n = E_1 + E_{given} = -13.6 \, eV + 12.4 \, eV = -1.2 \, eV$.The energy of an electron in the $n^{th}$ orbit is given by $E_n = -\frac{13.6}{n^2} \, eV$.Equating the two expressions:$-\frac{13.6}{n^2} = -1.2$$n^2 = \frac{13.6}{1.2} \approx 11.33$.Since $n$ must be an integer,we look for the nearest orbit. The electron will transition to the $n = 3$ orbit $(E_3 = -1.51 \, eV)$ as it does not have enough energy to reach the $n = 4$ orbit $(E_4 = -0.85 \, eV)$.

When an electron in the ground state of a hydrogen atom is given $12.4 \, eV$ of energy,to which orbit will it transition?

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