An electronic transition in a hydrogen atom r | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The energy of an electron in the $n^{th}$ orbit is given by $E_n = -\frac{313.52 Z^2}{n^2} \ kcal \ mol^{-1}$.For a hydrogen atom, $Z = 1$.The Lym

Vedclass · Accepted Answer

$-313.6, -78.4$

Vedclass · Answer

(B) The energy of an electron in the $n^{th}$ orbit is given by $E_n = -\frac{313.52 Z^2}{n^2} \ kcal \ mol^{-1}$.For a hydrogen atom, $Z = 1$.The Lyman series corresponds to transitions ending at $n_1 = 1$.The $H_\alpha$ line in the Lyman series corresponds to the transition from $n_2 = 2$ to $n_1 = 1$.Energy in the $n_1 = 1$ orbit: $E_1 = -\frac{313.52 	imes (1)^2}{(1)^2} = -313.52 \ kcal \ mol^{-1} \approx -313.6 \ kcal \ mol^{-1}$.Energy in the $n_2 = 2$ orbit: $E_2 = -\frac{313.52 	imes (1)^2}{(2)^2} = -\frac{313.52}{4} = -78.38 \ kcal \ mol^{-1} \approx -78.4 \ kcal \ mol^{-1}$.Thus, the energies are $-313.6 \ kcal \ mol^{-1}$ and $-78.4 \ kcal \ mol^{-1}$.

An electronic transition in a hydrogen atom results in the formation of the $H_\alpha$ line of the hydrogen spectrum in the Lyman series. The energies associated with the electron in each of the orbits involved in the transition (in $kcal \ mol^{-1}$) are:

Similar Questions

If the radius of the hydrogen atom is $53 \, pm$, the radius of the $He^{+}$ ion is closest to $...... \, pm$.

Which of the following electronic transitions in a hydrogen atom will require the highest energy?

The observation of the line spectrum of the hydrogen atom indicates that

For a $H$-atom,what is the wavelength in $nm$ associated with the transition of an electron from an infinite energy level to the first energy level? (Rydberg constant = $1.097 \times 10^{7} \ m^{-1}$)

Which of the following is the radius of the first orbit of $He^{+}$ (in $pm$)?

Vedclass Test Series

Exam Paper Generator

Online Exam Module