An electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let $\lambda_n, \lambda_g$ be the de Broglie wavelength of the electron in the $n^{th}$ state and the ground state respectively. Let $\Lambda_n$ be the wavelength of the emitted photon in the transition from the $n^{th}$ state to the ground state. For large $n,$ ($A, B$ are constants)

Question

The nuclides of $^{197}_{79}\mathrm{Au}$ and $^{196}_{80}\mathrm{Hg}$ are called ______ of each other

Answer 1

To solve this problem, we need to analyze the given situation regarding the de Broglie wavelength of an electron and the wavelength of the emitted photon when the electron transitions from the $ n^{th} $ excited state to the ground state in a hydrogen atom.

Understanding De Broglie Wavelength: The de Broglie wavelength $\lambda_n$ of an electron in the $ n^{th} $ excited state can be expressed as: \[ \lambda_n = \frac{h}{p_n} \] where $ h $ is the Planck's constant and $ p_n $ is the momentum of the electron in the $ n^{th} $ state.
Energy Levels of Hydrogen Atom: The energy of an electron in the $ n^{th} $ state is given by: \[ E_n = - \frac{13.6 \, \text{eV}}{n^2} \] The energy difference between levels is emitted as a photon, having wavelength $\Lambda_n$.
Photon Wavelength Relation: The energy difference $ \Delta E $ during the transition from the $ n^{th} $ state to the ground state is: \[ \Delta E = E_1 - E_n = 13.6 \, \text{eV} \left( 1 - \frac{1}{n^2} \right) \] \Lambda_n = \frac{hc}{\Delta E} \] For large $ n $, $ \frac{1}{n^2} $ is very small, simplifying the energy difference.
Substitution and Approximation: As $ n \to \infty $, $ \Delta E \approx 13.6 \, \text{eV} $ for large $ n$. The de Broglie wavelength is related to kinetic energy $ E_k $ where: \[ \frac{1}{\lambda_n^2} \propto \frac{1}{p_n^2} \propto E_k \]
Conclusion with the Correct Option: Incorporating this into the expression for $\Lambda_n$, and noting the dependency on $\frac{1}{\lambda_n^2}$, we find that the approximation: \[ \Lambda_n \approx A + \frac{B}{\lambda_n^2} \] fits the expected behavior for large $ n $, where $ A $ and $ B $ are constants.

Each step above justifies the reasoning for choosing the correct option, eliminating any other possibilities based on the dependency of $\Lambda_n$ on $\lambda_n$.

Answer 2

To solve this problem, we need to analyze the given situation regarding the de Broglie wavelength of an electron and the wavelength of the emitted photon when the electron transitions from the $ n^{th} $ excited state to the ground state in a hydrogen atom.

Understanding De Broglie Wavelength: The de Broglie wavelength $\lambda_n$ of an electron in the $ n^{th} $ excited state can be expressed as: \[ \lambda_n = \frac{h}{p_n} \] where $ h $ is the Planck's constant and $ p_n $ is the momentum of the electron in the $ n^{th} $ state.
Energy Levels of Hydrogen Atom: The energy of an electron in the $ n^{th} $ state is given by: \[ E_n = - \frac{13.6 \, \text{eV}}{n^2} \] The energy difference between levels is emitted as a photon, having wavelength $\Lambda_n$.
Photon Wavelength Relation: The energy difference $ \Delta E $ during the transition from the $ n^{th} $ state to the ground state is: \[ \Delta E = E_1 - E_n = 13.6 \, \text{eV} \left( 1 - \frac{1}{n^2} \right) \] \Lambda_n = \frac{hc}{\Delta E} \] For large $ n $, $ \frac{1}{n^2} $ is very small, simplifying the energy difference.
Substitution and Approximation: As $ n \to \infty $, $ \Delta E \approx 13.6 \, \text{eV} $ for large $ n$. The de Broglie wavelength is related to kinetic energy $ E_k $ where: \[ \frac{1}{\lambda_n^2} \propto \frac{1}{p_n^2} \propto E_k \]
Conclusion with the Correct Option: Incorporating this into the expression for $\Lambda_n$, and noting the dependency on $\frac{1}{\lambda_n^2}$, we find that the approximation: \[ \Lambda_n \approx A + \frac{B}{\lambda_n^2} \] fits the expected behavior for large $ n $, where $ A $ and $ B $ are constants.

Each step above justifies the reasoning for choosing the correct option, eliminating any other possibilities based on the dependency of $\Lambda_n$ on $\lambda_n$.

The Correct Option is A

Solution and Explanation

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