Question

A hydrogen atom in its ground state is irradiated by light of wavelength \(1026\ \text{Å}\). The electron gets excited to higher energy state \(n\). Given the ground state energy is \(-13.6\ \text{eV}\) and \(hc=1242\ \text{eV}\cdot \text{nm}\), then the value of principal quantum number \(n\) is

• A. \(5\)
• B. \(3\)
• C. \(4\)
• D. \(2\)

Hint:

Find the photon energy from hc over wavelength, add it to the ground state energy and match with E = -13.6 / n squared.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
The absorbed photon energy raises the electron from \(n=1\) to a higher level \(n\). Energy levels are \(E_{n}=-\dfrac{13.6}{n^{2}}\ \text{eV}\).

Step 2: Photon energy
\(\lambda=1026\ \text{Å}=102.6\ \text{nm}\)
\(E=\dfrac{1242}{102.6}\approx 12.1\ \text{eV}\)

Step 3: Energy of excited level
\(E_{n}=-13.6+12.1=-1.5\ \text{eV}\)

Step 4: Find n
\(-\dfrac{13.6}{n^{2}}=-1.5\Rightarrow n^{2}\approx 9\Rightarrow n=3\)

Final Answer:
The electron is excited to the level n equal to 3. \[ \boxed{\text{(B) }3} \]