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Question

(C) The energy change for a transition in a hydrogen atom is given by the formula: $\Delta E = 2.18 	imes 10^{-18} \ J 	imes (\frac{1}{n_1^2} - \fra

Vedclass · Accepted Answer

$3.03 	imes 10^{-19} \ J \ atom^{-1}$

Vedclass · Answer

(C) The energy change for a transition in a hydrogen atom is given by the formula: $\Delta E = 2.18 	imes 10^{-18} \ J 	imes (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.Here,$n_1 = 2$ and $n_2 = 3$.$\Delta E = 2.18 	imes 10^{-18} 	imes (\frac{1}{2^2} - \frac{1}{3^2})$$\Delta E = 2.18 	imes 10^{-18} 	imes (\frac{1}{4} - \frac{1}{9})$$\Delta E = 2.18 	imes 10^{-18} 	imes (\frac{9-4}{36})$$\Delta E = 2.18 	imes 10^{-18} 	imes \frac{5}{36}$$\Delta E = 0.3027 	imes 10^{-18} \ J \ atom^{-1}$$\Delta E = 3.03 	imes 10^{-19} \ J \ atom^{-1}$.

Calculate the energy released when an electron in a hydrogen atom transitions from the $3^{rd}$ energy level to the $2^{nd}$ energy level.

Similar Questions

The energy of the electron at infinite distance from the nucleus in Bohr’s model is taken as:

If the radius of the electron orbit in the excited state of a hydrogen atom is $476.1 \ pm$, the energy of the electron in that excited state in $J$ is (Radius and energy of the electron in the first orbit of a hydrogen atom are $52.9 \ pm$ and $-2.18 \times 10^{-18} \ J$ respectively).

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$A$: Thomson's atomic model is known as the plum pudding model.
$R$: The atom is considered as a positively charged sphere in which electrons are embedded.

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