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Consider two atoms: Hydrogen (Z = 1) and Carbon (Z = 6). If the Bohr radius of ground state electron orbit in hydrogen atom be R, what would be the Bohr radius of the single ground-state electron in the carbon atom? 6R R6 R/62 R/6

Consider two atoms: Hydrogen (Z = 1) and Carbon (Z = 6). If the Bohr radius of ground state electron orbit in hydrogen atom be R, what would be the Bohr radius of the single ground-state electron in the carbon atom? 6R R6 R/62 R/6

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### Bohr Radius in Hydrogen and Carbon Atoms **Problem Statement:** Consider two atoms: Hydrogen (Z = 1) and Carbon (Z = 6). If the Bohr radius of the ground state electron orbit in a hydrogen atom is \( R \), what would be the Bohr radius of the single ground-state electron in the carbon atom? **Multiple Choice Options:** - \( 6R \) - \( R^6 \) - \( \frac{R}{6^2} \) - \( \frac{R}{6} \) To understand the relationship between the Bohr radius of different atoms, we use the formula for the Bohr radius of a hydrogen-like atom: \[ r_n = \frac{n^2 \hbar^2}{Z k e^2 m_e} \] Where: - \( r_n \) is the radius of the orbit. - \( n \) is the principal quantum number. - \( \hbar \) is the reduced Planck's constant. - \( e \) is the electron charge. - \( k \) is Coulomb's constant. - \( m_e \) is the electron mass. - \( Z \) is the atomic number. For the ground state (\( n = 1 \)): \[ r = \frac{\hbar^2}{Z k e^2 m_e} \] Given that for hydrogen (Z = 1), the Bohr radius \( R \) is: \[ R = \frac{\hbar^2}{k e^2 m_e} \] For the carbon atom (\( Z = 6 \)): \[ r_{carbon} = \frac{\hbar^2}{6 k e^2 m_e} = \frac{R}{6} \] Therefore, the correct answer is: - \( \frac{R}{6} \) This question tests the understanding of the Bohr model and how the atomic number \( Z \) affects the radius of the electron orbit in different atoms.
Transcribed Image Text:### Bohr Radius in Hydrogen and Carbon Atoms **Problem Statement:** Consider two atoms: Hydrogen (Z = 1) and Carbon (Z = 6). If the Bohr radius of the ground state electron orbit in a hydrogen atom is \( R \), what would be the Bohr radius of the single ground-state electron in the carbon atom? **Multiple Choice Options:** - \( 6R \) - \( R^6 \) - \( \frac{R}{6^2} \) - \( \frac{R}{6} \) To understand the relationship between the Bohr radius of different atoms, we use the formula for the Bohr radius of a hydrogen-like atom: \[ r_n = \frac{n^2 \hbar^2}{Z k e^2 m_e} \] Where: - \( r_n \) is the radius of the orbit. - \( n \) is the principal quantum number. - \( \hbar \) is the reduced Planck's constant. - \( e \) is the electron charge. - \( k \) is Coulomb's constant. - \( m_e \) is the electron mass. - \( Z \) is the atomic number. For the ground state (\( n = 1 \)): \[ r = \frac{\hbar^2}{Z k e^2 m_e} \] Given that for hydrogen (Z = 1), the Bohr radius \( R \) is: \[ R = \frac{\hbar^2}{k e^2 m_e} \] For the carbon atom (\( Z = 6 \)): \[ r_{carbon} = \frac{\hbar^2}{6 k e^2 m_e} = \frac{R}{6} \] Therefore, the correct answer is: - \( \frac{R}{6} \) This question tests the understanding of the Bohr model and how the atomic number \( Z \) affects the radius of the electron orbit in different atoms.
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