An electron in the hydrogen atom initially in the fourth excited state makes a transition to $ n^{th} $ energy state by emitting a photon of energy 2.86 eV. The integer value of n will be 1cm.

Question

$Li^{+2} \longrightarrow Li^{+3} + e^-$ :
Energy required for this process.
Given ionisation energy for ground state of hydrogen atom is $2.17 \times 10^{-18} \text{ J}$.

Answer 1

This task involves determining the final principal quantum number ($n$) of an electron in a hydrogen atom following a transition from a higher energy level, accompanied by the emission of a photon with a specified energy.

Concept Used:

The energy of an electron within the $k^{th}$ energy level (or orbit) of a hydrogen atom is defined by the following equation:

\[ E_k = -\frac{13.6}{k^2} \text{ eV} \]

The "ground state" is designated by $k=1$. The "first excited state" is at $k=2$, the "second excited state" at $k=3$, and so forth. Consequently, the $m^{th}$ excited state corresponds to the principal quantum number $k = m+1$.

Upon transitioning from a higher initial energy state $E_i$ (characterized by quantum number $k_i$) to a lower final energy state $E_f$ (characterized by quantum number $k_f$), an electron emits a photon. The energy of this emitted photon ($\Delta E$) is equivalent to the difference between these energy levels.

\[ \Delta E = E_i - E_f \]

Step-by-Step Solution:

Step 1: Identify the initial principal quantum number ($k_i$) of the electron.

The problem specifies that the electron begins in the "fourth excited state". Applying the established relationship where the $m^{th}$ excited state corresponds to $k = m+1$, for the fourth excited state ($m=4$), the principal quantum number is calculated as:

\[ k_i = 4 + 1 = 5 \]

Step 2: Compute the energy of the electron in its initial state ($E_i$).

Using the energy formula with the initial quantum number $k_i = 5$:

\[ E_i = E_5 = -\frac{13.6}{5^2} \text{ eV} = -\frac{13.6}{25} \text{ eV} \] \[ E_i = -0.544 \text{ eV} \]

Step 3: Utilize the energy of the emitted photon to determine the energy of the final state ($E_n$).

The energy of the emitted photon is provided as $\Delta E = 2.86$ eV. The electron transitions to the $n^{th}$ energy state, with energy $E_n$. The governing equation is:

\[ \Delta E = E_i - E_n \]

To solve for $E_n$, the equation is rearranged as follows:

\[ E_n = E_i - \Delta E \]

Substituting the known values yields:

\[ E_n = -0.544 \text{ eV} - 2.86 \text{ eV} \] \[ E_n = -3.404 \text{ eV} \]

Step 4: Ascertain the principal quantum number ($n$) for the final energy state.

The energy formula is applied once more, this time using the final energy $E_n$:

\[ E_n = -\frac{13.6}{n^2} \text{ eV} \]

Inserting the calculated value of $E_n$:

\[ -3.404 = -\frac{13.6}{n^2} \]

Solving for $n^2$:

\[ n^2 = \frac{13.6}{3.404} \] \[ n^2 \approx 3.995 \]

Given that the principal quantum number $n$ must be an integer, $n^2$ is rounded to the nearest integer.

\[ n^2 = 4 \] \[ n = \sqrt{4} = 2 \]

Therefore, the integer value of n corresponding to the final energy state is 2.

An electron in the hydrogen atom initially in the fourth excited state makes a transition to \( n^{th} \) energy state by emitting a photon of energy 2.86 eV. The integer value of n will be 1cm.

Show Hint

Correct Answer: 2

Solution and Explanation

Concept Used:

Step-by-Step Solution:

Top Questions on Atomic Physics

Questions Asked in JEE Main exam