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 $n$?
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 $n$?
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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.
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.
Updated On: Jun 16, 2026
- $\sqrt{\frac{E_1}{E_1 + E_a}}$
- $\sqrt{\frac{E_1}{E_1 - E_a}}$
- $\sqrt{\frac{E_a}{E_1 - E_a}}$
- $\sqrt{\frac{E_a}{E_1 + E_a}}$
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Question:
This question relates the absorption of a photon by a ground-state hydrogen electron to its subsequent transition to a higher principal quantum number $n$.
Step 2: Key Formulas and Approach:
1. Energy of an electron in the $n$-th state of a hydrogen-like atom:
\[ E_n = \frac{E_1}{n^2} \]
where $E_1$ is the ground state energy.
2. Energy conservation for photon absorption:
\[ E_n = E_1 + E_a \]
Step 3: Detailed Explanation:
• The energy level of a hydrogen atom is quantized and given by:
\[ E_n = \frac{E_1}{n^2} \]
where $E_1 \approx -13.6\text{ eV}$ represents the negative ground-state energy, and $n$ is the principal quantum number.
• The electron initially in the ground state has energy $E_1$.
• Upon absorbing a photon of energy $E_a$, the final energy of the electron becomes:
\[ E_n = E_1 + E_a \]
• Equating this to the formula for $E_n$:
\[ \frac{E_1}{n^2} = E_1 + E_a \]
• Solve for $n^2$ by taking the reciprocal:
\[ n^2 = \frac{E_1}{E_1 + E_a} \]
• Taking the square root on both sides:
\[ n = \sqrt{\frac{E_1}{E_1 + E_a}} \]
• Since both $E_1$ and $E_1 + E_a$ are negative quantities for a bound state, their ratio is positive, giving a real and valid principal quantum number $n$.
Step 4: Final Answer:
The value of $n$ is $\sqrt{\frac{E_1}{E_1 + E_a}}$, which corresponds to Option (A).
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