Question

If an electron in \(n=4\) orbit of hydrogen atom jumps to \(n=3\), the energy released and wavelength emitted are:

• A. \(0.66\,eV , 1.88 \times 10^{-6}\,m\)
• B. \(1.89\,eV , 1.98 \times 10^{-7}\,m\)
• C. \(0.29\,eV , 1.78 \times 10^{-5}\,m\)
• D. \(0.98\,eV , 0.93 \times 10^{-6}\,m\)

Hint:

For hydrogen transitions: \[ \lambda(\text{nm}) = \frac{1240}{E(\text{eV})} \] is the fastest way to calculate wavelength.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
When an electron jumps from a higher energy level (\(n_{2}\)) to a lower energy level (\(n_{1}\)), energy is released in the form of a photon. The energy corresponds to the difference between the two levels.
Step 2: Key Formula or Approach:
Energy levels of H-atom: \(E_{n} = -\frac{13.6}{n^{2}}\text{ eV}\).
Wavelength: \(\lambda = \frac{hc}{\Delta E}\) or \(\frac{1}{\lambda} = R \left( \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right)\).
: Detailed Explanation:
1. Calculate Energy Released (\(\Delta E\)):
\[ E_{4} = -\frac{13.6}{4^{2}} = -0.85\text{ eV} \]
\[ E_{3} = -\frac{13.6}{3^{2}} = -1.51\text{ eV} \]
\[ \Delta E = E_{4} - E_{3} = -0.85 - (-1.51) = 0.66\text{ eV} \]
2. Calculate Wavelength (\(\lambda\)):
Using \(\lambda = \frac{12400}{\Delta E\text{ (in eV)}}\text{ \r{A}}\):
\[ \lambda = \frac{12400}{0.66} \approx 18787\text{ \r{A}} = 1.8787 \times 10^{-6}\text{ m} \]
Comparing with options, \(1.88 \times 10^{-6}\text{ m}\) is the closest match.
Step 3: Final Answer:
The released energy is \(0.66\text{ eV}\) and wavelength is \(1.88 \times 10^{-6}\text{ m}\).