Velocity of an Electron in the First Excited State
In the Bohr model of the hydrogen atom, the first excited state corresponds to the principal quantum number \( n = 2 \). Below is the derivation of the electron's velocity in this state.
Derivation
The angular momentum \( L \) of the electron in the \( n \)-th orbit is quantized as:
\[ L = m_e v_n r_n = n \hbar, \]where \( m_e \) is the electron mass, \( v_n \) is the velocity, \( r_n \) is the orbital radius, \( n \) is the principal quantum number, and \( \hbar = \frac{h}{2\pi} \) is the reduced Planck's constant.
The orbital radius in the Bohr model is:
\[ r_n = n^2 a_0, \]where \( a_0 = \frac{4\pi \epsilon_0 \hbar^2}{m_e e^2} \) is the Bohr radius, \( \epsilon_0 \) is the vacuum permittivity, and \( e \) is the elementary charge.
Substitute \( r_n \) into the angular momentum equation:
\[ m_e v_n (n^2 a_0) = n \hbar \implies v_n = \frac{n \hbar}{m_e n^2 a_0} = \frac{\hbar}{m_e n a_0}. \]Substitute the expression for \( a_0 \):
\[ v_n = \frac{\hbar}{m_e n \left( \frac{4\pi \epsilon_0 \hbar^2}{m_e e^2} \right)} = \frac{\hbar \cdot m_e e^2}{m_e n \cdot 4\pi \epsilon_0 \hbar^2} = \frac{e^2}{4\pi \epsilon_0 n \hbar}. \]Alternatively, using the fine structure constant \( \alpha = \frac{e^2}{4\pi \epsilon_0 \hbar c} \), where \( c \) is the speed of light:
\[ v_n = \frac{\alpha c}{n}. \]For the first excited state (\( n = 2 \)):
\[ v_2 = \frac{e^2}{8\pi \epsilon_0 \hbar} = \frac{\alpha c}{2}. \]This is the velocity of the electron in the first excited state of the hydrogen atom in the Bohr model.