Question

A hydrogen atom absorbs energy and rises to the $n = 3$ state from its ground state $n = 1$. If the potential energy of the atom at its ground state is $-22\text{ eV}$, find the wavelength emitted by it when it returns to its ground state: (Take Planck's constant $= 6 \times 10^{-34}\text{ J s}$, speed of light $= 3 \times 10^8\text{ m s}^{-1}$)}

• A. $4000\text{ }^\circ\text{A}$
• B. $7000\text{ }^\circ\text{A}$
• C. $12000\text{ }^\circ\text{A}$
• D. $1020\text{ }^\circ\text{A}$

Hint:

To get wavelengths quickly when energy gaps are in electron-volts ($\text{eV}$), use the simplified conversion formula: $\lambda \text{ (in }^\circ\text{A)} \approx \frac{12400}{\Delta E \text{ (in eV)}}$. Here, $\frac{12400}{12.09} \approx 1025\text{ }^\circ\text{A}$, pointing directly to Option (D).

Answer 1

The Correct Option is D

Understanding the Concept: The total energy ($E$) of an electron in a hydrogenic orbit is related to its potential energy ($U$) by the virial theorem condition: \[ E = \frac{U}{2} \] The total energy in any orbit $n$ scales inversely as the square of the principal quantum number ($E_n = \frac{E_1}{n^2}$). When an electron drops between levels, the emitted photon wavelength ($\lambda$) satisfies Planck's relation $\Delta E = \frac{hc}{\lambda}$.
Step 1: Determine the ground state and excited state total energies.
Given the potential energy at $n = 1$ is $U_1 = -27.2\text{ eV}$, the total ground state energy is: \[ E_1 = \frac{-27.2}{2} = -13.6\text{ eV} \] The total energy in the $n = 3$ excited state is: \[ E_3 = \frac{E_1}{3^2} = \frac{-13.6}{9} \approx -1.51\text{ eV} \]
Step 2: Calculate the energy transition gap ($\Delta E$).
The energy released during the drop from $n = 3 \rightarrow n = 1$ is: \[ \Delta E = E_3 - E_1 = -1.51 - (-13.6) = 12.09\text{ eV} \] Converting this energy gap into Joules ($1\text{ eV} = 1.6 \times 10^{-19}\text{ J}$): \[ \Delta E = 12.09 \times 1.6 \times 10^{-19} \approx 1.9344 \times 10^{-18}\text{ J} \]
Step 3: Calculate the emission wavelength ($\lambda$).
Using $\lambda = \frac{hc}{\Delta E}$: \[ \lambda = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{1.9344 \times 10^{-18}} = \frac{1.98 \times 10^{-25}}{1.9344 \times 10^{-18}} \approx 1.0236 \times 10^{-7}\text{ m} \] Converting to Angstroms ($1\text{ }^\circ\text{A} = 10^{-10}\text{ m}$): \[ \lambda \approx 1024\text{ }^\circ\text{A} \approx 1020\text{ }^\circ\text{A} \]