If the potential energy of a hydrogen atom in | vedclass

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

Question

(B) The potential energy $U$ of an electron in a hydrogen atom at a state $n$ is given by $U_n = -27.2 / n^2 \ eV$.For the first excited state $(n = 2

Vedclass · Accepted Answer

$6.8 eV$

Vedclass · Answer

(B) The potential energy $U$ of an electron in a hydrogen atom at a state $n$ is given by $U_n = -27.2 / n^2 \ eV$.For the first excited state $(n = 2)$, the potential energy is $U_2 = -27.2 / 2^2 = -6.8 \ eV$.According to the problem, we assume $U_2 = 0$. This implies we are adding a constant $C = 6.8 \ eV$ to the potential energy scale.The total energy $E_n$ of a hydrogen atom is given by $E_n = -13.6 / n^2 \ eV$.For $n = \infty$, the standard total energy is $E_{\infty} = 0 \ eV$.Since we shifted the potential energy scale by adding $6.8 \ eV$, the new total energy $E'_{\infty}$ will be $E_{\infty} + 6.8 \ eV = 0 + 6.8 \ eV = 6.8 \ eV$.

If the potential energy of a hydrogen atom in the first excited state is assumed to be zero, then the total energy of the $n = \infty$ state is,

Similar Questions

The ground state energy of a hydrogen atom is $-13.6 \ eV$. The potential energy of the electron in the first excited state of hydrogen is (in $eV$)

How much energy is needed to release the electron from the second excited state in the hydrogen atom (in $eV$)?

The ground state energy of the hydrogen atom is $-13.6 \ eV$. The kinetic and potential energy of the electron in the second excited state are,respectively:

In the above figure,$D$ and $E$ respectively represent:

The binding energy of an electron in a $H$-atom with $n = 4$ is ....... $eV$. (Magnitude only)

Vedclass Test Series

Exam Paper Generator

Online Exam Module