With the increase in principal quantum number | 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 of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} 	ext{ eV}$.The energy difference between

Vedclass · Accepted Answer

Decreases

Vedclass · Answer

(B) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} 	ext{ eV}$.The energy difference between two successive levels $n$ and $n+1$ is $\Delta E = E_{n+1} - E_n = -13.6 \left( \frac{1}{(n+1)^2} - \frac{1}{n^2} ight) = 13.6 \left( \frac{1}{n^2} - \frac{1}{(n+1)^2} ight) = 13.6 \left( \frac{(n+1)^2 - n^2}{n^2(n+1)^2} ight) = 13.6 \left( \frac{2n+1}{n^2(n+1)^2} ight)$.As the principal quantum number $n$ increases,the denominator $n^2(n+1)^2$ increases much faster than the numerator $(2n+1)$.Therefore,the energy difference $\Delta E$ decreases as $n$ increases.From the provided image,we can see that the energy gap between levels decreases as we move to higher energy states: $E_1 > E_2 > E_3$.

With the increase in principal quantum number,the energy difference between two successive energy levels

Similar Questions

An electron in a hydrogen atom first jumps from the third excited state to the second excited state and then from the second excited state to the first excited state. The ratio of the wavelengths $\lambda_1 : \lambda_2$ emitted in the two cases is

If $\lambda_1$ and $\lambda_2$ are the wavelengths of the photons emitted when electrons in the $n^{\text{th}}$ orbit of a hydrogen atom fall to the first excited state and ground state respectively,then the value of $n$ is:

$A$ hydrogen atom (ionisation potential $13.6 \ eV$) makes a transition from the third excited state to the first excited state. The energy of the photon emitted in the process is.......$eV$.

An electron of a stationary hydrogen atom transits from the fourth energy level to the ground level. The velocity of the hydrogen atom acquired as a result of photon emission will be:
{ $R =$ Rydberg constant,$m =$ mass of hydrogen atom }

In a hydrogen atom,the electron is in the $n^{th}$ excited state. It may come down to the second excited state by emitting ten different wavelengths. What is the value of $n$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module