With increasing principal quantum number,the | vedclass

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

Question

(A) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by $E_n = -13.6 	imes \frac{Z^2}{n^2} 	ext{ eV}$.The energy dif

Vedclass · Accepted Answer

decreases

Vedclass · Answer

(A) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by $E_n = -13.6 	imes \frac{Z^2}{n^2} 	ext{ eV}$.The energy difference between adjacent levels $n$ and $n+1$ is $\Delta E = E_{n+1} - E_n = 13.6 	imes Z^2 \left( \frac{1}{n^2} - \frac{1}{(n+1)^2} ight) = 13.6 	imes Z^2 \left( \frac{2n+1}{n^2(n+1)^2} ight)$.As $n$ increases,the term $\frac{2n+1}{n^2(n+1)^2}$ decreases rapidly.For example:For $n=1$ to $2$,$\Delta E = 13.6 	imes (1 - 0.25) = 10.2 	ext{ eV}$.For $n=2$ to $3$,$\Delta E = 13.6 	imes (0.25 - 0.111) = 1.89 	ext{ eV}$.Thus,the energy difference decreases as the principal quantum number increases.

With increasing principal quantum number,the energy difference between adjacent energy levels in $H$ atom

Similar Questions

The limiting line in the Balmer series of the $H$ spectrum has a frequency of ....... $(R_\infty = 3.29 \times 10^{15} \, cycles/s)$

The electrostatic force of attraction between the electron and the nucleus in a hydrogen atom is given by:

What is the value of the frequency of radiation when a transition occurs between two stationary states that differ in energy by $\Delta E$?

If the wavelength of the first line in the Balmer series is $656 \ nm$,then the wavelength of its second line and limiting line respectively are

If an electron transitions from $n = 5$ to $n = 2$ in a hydrogen spectrum,identify the series and the region of the spectrum.

Vedclass Test Series

Exam Paper Generator

Online Exam Module