Question

An electron in the ground state (with energy $E_1$) of a hydrogen atom, absorbs a photon of energy $E_a$, and gets excited to a higher energy level of principal quantum number $n$. What is the value of $n$?

• A. $\sqrt{\frac{E_1}{E_1 + E_a}}$
• B. $\sqrt{\frac{E_1}{E_1 - E_a}}$
• C. $\sqrt{\frac{E_a}{E_1 - E_a}}$
• D. $\sqrt{\frac{E_a}{E_1 + E_a}}$

Hint:

Always check the dimensions and sign of the terms.
Since $n$ must be a positive integer greater than 1, and $E_1$ is negative, $E_1 + E_a$ is less negative (closer to zero).
Thus, $\frac{E_1}{E_1 + E_a} > 1$, which is required for $n > 1$.
This sign analysis helps you eliminate incorrect options instantly.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
According to Bohr's theory of the hydrogen atom, the total energy of an electron in any given stationary orbit is quantized.
When an electron transitions from a lower energy state to a higher energy state, it absorbs a photon whose energy is exactly equal to the difference in energy between those two quantum levels.
Step 2: Key Formula or Approach:
The total energy of an electron in the \( n^{\text{th}} \) Bohr orbit of a hydrogen atom scales inversely with the square of the principal quantum number:
\[ E_n = \frac{E_1}{n^2} \]
where \( E_1 \) represents the total potential and kinetic energy of the electron in its ground state (\( n = 1 \)). Note that in standard atomic systems, the numerical value of \( E_1 \) is natively negative (e.g., \( -13.6 \text{ eV} \)).
The energy of the absorbed photon (\( E_a \)) required to move the electron from the ground state to a higher state \( n \) is:
\[ E_a = E_n - E_1 \]
Step 3: Detailed Explanation:
Let us substitute the relationship for \( E_n \) into our energy balance equation:
\[ E_a = \frac{E_1}{n^2} - E_1 \]
We need to rearrange this expression to isolate the variable \( n \). First, add \( E_1 \) to both sides of the equation:
\[ E_a + E_1 = \frac{E_1}{n^2} \]
Next, take the reciprocal of both sides or rearrange the terms cross-multiplicatively to solve for \( n^2 \):
\[ n^2 = \frac{E_1}{E_a + E_1} \]
\[ n^2 = \frac{E_1}{E_1 + E_a} \]
Self-Correction and Sign Convention Analysis:
Because \( E_1 \) is structurally a negative quantity (representing a bound state), working with its absolute values or treating \( E_1 \) as a negative symbol alters the structural look of the choices. Let us re-verify how the choices are written:
If we treat \( E_1 \) as its explicit negative standard value (where \( E_1 = -R_{\text{H}} \)), let us re-write the relation to ensure the denominator stays positive and algebraically clean:
\[ E_a = \left(-\frac{R_{\text{H}}}{n^2}\right) - (-R_{\text{H}}) \]
\[ E_a = R_{\text{H}} - \frac{R_{\text{H}}}{n^2} \]
\[ \frac{R_{\text{H}}}{n^2} = R_{\text{H}} - E_a \]
\[ n^2 = \frac{R_{\text{H}}}{R_{\text{H}} - E_a} \]
Replacing \( R_{\text{H}} \) with \( -E_1 \) (since \( E_1 = -R_{\text{H}} \)):
\[ n^2 = \frac{-E_1}{-E_1 - E_a} \]
Multiply both the numerator and the denominator by \(-1\):
\[ n^2 = \frac{E_1}{E_1 - E_a} \]
Taking the square root of both sides gives:
\[ n = \sqrt{\frac{E_1}{E_1 - E_a}} \]
This matches Option (B) perfectly under standard thermodynamic sign configurations.
Step 4: Final Answer:
The principal quantum number value of \( n \) is \(\sqrt{\frac{E_1}{E_1 - E_a}}\).