Question

According to the Bohr model of the hydrogen atom, what is the magnitude of the translational angular momentum of the electron (relative to the location of the proton) when the atom is in the 2^nd excited state above the ground state$\mathbf{(}\mathbf{N}\mathbf{=}\mathbf{3}\mathbf{)}\mathbf{?}$

Short answer

The magnitude of the translational angular momentum of the electron is $3.162\times {10}^{-34}\text{J}·\text{s}$.

Step-by-step solution

Definition of angular momentum

The angular momentum is defined as the multiplication of the moment of inertia and the angular velocity of a body.

Find the magnitude of the translational angular momentum of the electron.

The possible state of the hydrogen atom whose translation angular momentum is an integer multiple of$h$ .

$\left|{\overrightarrow{L}}_{trans,C}\right|=Nh······\left(1\right)$

Here$h\to$ Planck’s constant$\left(1.054\times {10}^{-34}\text{J}·\text{s}\right)$.

For the second excited state of the atom,$N=3$

On substituting all numerical values in equation(1), the magnitude of the translational angular momentum of the electron as

$\begin{array}{rcl}\left|{\overrightarrow{L}}_{trans,C}\right|& =& Nh\\ & =& \left(3\right)\left(1.054\times {10}^{-34}\text{J}·\text{s}\right)\\ & =& 3.162\times {10}^{-34}\text{J}·\text{s}\\ & & \end{array}$

Hence, the magnitude of the translational angular momentum of the electron is $3.162\times {10}^{-34}\text{J}·\text{s}$.