For He\(^+\), a transition takes place from the orbit of radius 105.8 pm to the orbit of radius 26.45 pm. The wavelength (in nm) of the emitted photon during the transition is _________.
Use the following constants:
- Bohr radius, \( a_0 = 52.9 \text{ pm} \)
- Rydberg constant, \( R_H = 2.2 \times 10^{-18} \text{ J} \)
- Planck's constant, \( h = 6.6 \times 10^{-34} \text{ J s} \)
- Speed of light, \( c = 3 \times 10^8 \text{ m s}^{-1} \)
Detailed Step-by-Step Solution
This problem requires us to first find the principal quantum numbers (\(n\)) corresponding to the given radii, calculate the energy difference, and then find the wavelength.
Step 1: Find the Principal Quantum Numbers (\(n_1\) and \(n_2\))
The radius of the \(n^{\text{th}}\) Bohr orbit for a hydrogen-like species is: \( r_n = \frac{n^2}{Z} \times a_0 \)
For He\(^+\), the atomic number \( Z = 2 \).
Initial Orbit (\(n_i\)):
\( n_i^2 = \frac{105.8 \times 2}{52.9} = \frac{211.6}{52.9} = 4 \)
\( \mathbf{n_i = 2} \)
Final Orbit (\(n_f\)):
\( n_f^2 = \frac{26.45 \times 2}{52.9} = \frac{52.9}{52.9} = 1 \)
\( \mathbf{n_f = 1} \)
So, the transition is from \( n = 2 \) to \( n = 1 \).
Step 2: Calculate the Energy of the Emitted Photon (\(\Delta E\))
The formula for the energy difference during a transition is:
Substitute the known values (\( R_H = 2.2 \times 10^{-18} \text{ J}, Z = 2, n_f = 1, n_i = 2 \)):
\( \Delta E = (2.2 \times 10^{-18}) \times 4 \times \left( 1 - \frac{1}{4} \right) \)
\( \Delta E = (2.2 \times 10^{-18}) \times 4 \times \frac{3}{4} \)
\( \Delta E = 2.2 \times 10^{-18} \times 3 = \mathbf{6.6 \times 10^{-18} \text{ J}} \)
Step 3: Calculate the Wavelength (\(\lambda\))
Relate energy to wavelength using Planck's equation: \( \Delta E = \frac{hc}{\lambda} \implies \lambda = \frac{hc}{\Delta E} \)
Notice how elegantly the \( 6.6 \) values cancel out (this is typical of JEE Advanced making calculations simple if you follow the correct path):
\( \lambda = 3 \times 10^{-34 + 8 + 18} = 3 \times 10^{-8} \text{ m} \)
Convert the wavelength from meters to nanometers (\(1 \text{ nm} = 10^{-9} \text{ m}\)):
Conclusion: The wavelength of the emitted photon is exactly 30 nm. Therefore, the final integer answer is 30.
Navigating JEE Advanced Numericals
JEE Advanced questions often require you to work backward. Instead of giving you the orbit transitions (\(n_1\) and \(n_2\)), they give you physical observables like orbit radii. A crucial shortcut in this problem is recognizing that the given constants (\(6.6\) and \(2.2\)) are designed to cancel out perfectly if set up as a single equation before multiplying.
To master multi-step problem solving involving Bohr's Model, Rydberg's constant, and spectral transitions, study our comprehensive guide on the Structure of Atom Class 11 Chemistry.
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