Problem 1: Calculate The Electronic Transitions Nf To Ni In ✓ Solved

Problem 1calculate The Electronic Transitions Nf To Ni In Hydrogen

Calculate the electronic transitions (nf to ni) in hydrogen which will lead to emission of visible light ( nm). Problem 2: (a) Calculate the deBroglie wavelength for an electron with a kinetic energy of 20 keV. (b) Calculate the deBroglie wavelength for a proton with a kinetic energy of 20 keV. (c) Calculate the kinetic energy of an electron having a deBroglie wavelength of 10.0 nm. Problem 3: Using the solutions to Schrodinger’s equation listed in the lecture notes, determine the distance at which you would expect to find the outermost electron in berylium. Problem 4: (a) The ionization energy of Ne is 21.6 eV. Using the equation associated with the Bohr atomic model, back-calculate the effective nuclear charge that the outer-most electron sees. (b) In a similar fashion back-calculate the effective nuclear charge the outermost electron sees for sodium. (c) Can you comment on the values of effective nuclear charge relative to the number of protons for each of these two elements?

Sample Paper For Above instruction

Introduction

Understanding electronic transitions in hydrogen and the related quantum mechanical concepts is fundamental to atomic physics. Analyzing the emission spectra, calculating de Broglie wavelengths for particles at high energies, and evaluating electron distributions within atoms through Schrödinger’s solutions are essential topics. This paper addresses these core concepts, with specific focus on hydrogen spectral transitions, de Broglie wavelengths, and effective nuclear charge in noble and alkali metals.

Electronic Transitions in Hydrogen Leading to Visible Light Emission

Hydrogen's spectral lines are governed by the Balmer series, which involves electronic transitions where the final energy level (n₁) is 2, and the initial energy levels (n₂) are higher integers (n₂ > 2). These transitions emit photons within the visible spectrum, approximately between 380 nm and 750 nm. To compute the specific transitions, we use the Rydberg formula:

\[

\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)

\]

where \( R \) is the Rydberg constant (approximately \( 1.097 \times 10^7 \) m\(^{-1}\)). For visible light, the wavelengths typically are associated with transitions from higher levels such as \( n_2 = 3, 4, 5 \) to \( n_1 = 2 \). Calculations indicate:

- \( n_2 = 3 \rightarrow 2 \): corresponds to the red line at about 656 nm.

- \( n_2 = 4 \rightarrow 2 \): produces the green-blue line near 486 nm.

- \( n_2 = 5 \rightarrow 2 \): results in a violet line around 434 nm.

Thus, the significant visible emissions originate mainly from these transitions, with the \( n_2 = 4 \rightarrow 2 \) transition being particularly prominent in the spectrum.

De Broglie Wavelength Calculations

(a) For an electron with kinetic energy \( KE = 20 \) keV, the de Broglie wavelength is given by:

\[

\lambda = \frac{h}{p}

\]

where \( p = \sqrt{2 m KE} \). Using relativistic corrections, the wavelength becomes:

\[

\lambda = \frac{h c}{\sqrt{(KE + m c^2)^2 - (m c^2)^2}}

\]

Substituting values:

- \( h \) (Planck’s constant) = \( 6.626 \times 10^{-34} \) Js,

- \( c \) (speed of light) = \( 3.0 \times 10^{8} \) m/s,

- \( m \) (electron rest mass) = \( 9.11 \times 10^{-31} \) kg,

- KE = 20,000 eV = \( 3.2 \times 10^{-15} \) J.

The relativistic formula yields a wavelength of approximately 2.5 pm, indicating significant relativistic effects at this energy.

(b) For a proton with the same kinetic energy, using similar principles:

- Proton mass \( m_p = 1.67 \times 10^{-27} \) kg,

the de Broglie wavelength is approximately 0.07 nm, larger than the electron's due to the larger mass.

(c) To find the electron's kinetic energy for a de Broglie wavelength of 10.0 nm, rearranging the de Broglie relation gives an energy of roughly 0.125 eV, showing that longer wavelengths correspond to very low kinetic energies.

Electron Distribution in Beryllium from Schrödinger’s Equation

Schrödinger’s solutions predict that the outermost electrons in beryllium (which has an atomic number of 4) are located at a certain average distance from the nucleus, associated with the 2s orbital. Using quantum mechanical calculations and known atomic radii, the outermost electrons are expected to reside approximately 0.9 Å (angstroms) from the nucleus, consistent with experimental measurements of atomic radii. This distribution is influenced by the effective nuclear charge experienced by the electrons, which modulates electron-electron repulsions and energy levels.

Effective Nuclear Charge in Neon and Sodium

(a) The ionization energy of neon (Ne) is 21.6 eV. Using the Bohr model, the effective nuclear charge \( Z_{eff} \) experienced by the outermost electrons can be approximated from:

\[

E_{ion} = \frac{Z_{eff}^2 R_H}{n^2}

\]

where \( R_H \) is the Rydberg energy (13.6 eV), and \( n \) is the principal quantum number. Rearranged, this yields \( Z_{eff} \approx 7.3 \), close to the actual nuclear charge minus shielding effects.

(b) For sodium (Na), with a first ionization energy of 5.14 eV, applying the same approach indicates \( Z_{eff} \) is roughly 1.0, reflecting the screening effect of inner electrons.

(c) Comparing these values, the effective nuclear charge for neon, near the number of protons (10), demonstrates minimal shielding, whereas sodium’s lower \( Z_{eff} \) indicates substantial electron shielding. These variations are consistent with expectations from atomic theory, illustrating how electron-electron interactions influence the effective nuclear charge felt by valence electrons.

Conclusion

The analysis of hydrogen spectral lines reveals that visible emissions predominantly result from specific electronic transitions between higher energy levels and the second energy level. The calculation of de Broglie wavelengths for electrons and protons demonstrates the wave-particle duality at high energies, with electrons exhibiting significantly shorter wavelengths due to their lower mass. Schrödinger’s solutions enable us to estimate electron locations, which align with experimental atomic radii. Finally, the comparison of effective nuclear charges in neon and sodium underscores the importance of electron shielding in determining atomic properties. These insights deepen our understanding of atomic structure, spectral phenomena, and quantum mechanical principles governing microscopic particles.

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