Question

In Bohr’s atomic model, an electron jumps from n=1 to n=3, how much energy will be absorbed?

Answer 1

Hint: when electron moves from lower stationary state to higher stationary state when required amount of energy is absorbed. The energy gap between two orbits is given by equation
$\Delta E={{R}_{H}}\left( \dfrac{1}{{{n}_{1}}^{2}}-\dfrac{1}{{{n}_{f}}^{2}} \right)$
Where ${{R}_{H}}$ is Rydberg constant.

Complete step by step answer:
-Bohr’s atomic model is based on following postulates:
-The electron in a hydrogen atom can move in a circular path around the nucleus. These paths are called orbits.
-These orbits are arranged around the nucleus concentrically.
-the energy of the electron in orbit does not change with time.
-when electron moves from lower stationary state to higher stationary state when required amount of energy is absorbed.
-when electron moves from higher stationary state to lower stationary state, energy is emitted.
-the energy change does not take place in a continuous manner.

-The frequency of radiation absorbed or emitted when transition occurs between two stationary states that differ in energy \[\Delta E\]is given by:
\[\nu =\dfrac{\Delta E}{h}=\dfrac{{{E}_{2}}-{{E}_{1}}}{h}\]
Where ${{E}_{1}}$ and ${{E}_{2}}$ are energies of lower and higher energy states.
-The stationary states for electrons are numbered n=1,2,3…

-The radii of stationary states are expressed as:
\[{{r}_{n}}={{n}^{2}}{{a}_{0}}\] where ${{a}_{0}}$=52.9pm. radius of first stationary state called bohr orbit.
-Energy associated with electron in its stationary state is given by
\[{{E}_{n}}=-{{R}_{H}}\dfrac{1}{{{n}^{2}}}\]
Where \[{{R}_{H}}\]is Rydberg constant and value for this constant is \[-2.18\times {{10}^{-18}}J\].

The energy of lowest state also called ground state is
${{E}_{1}}$=\[-2.18\times {{10}^{-18}}\dfrac{1}{{{(1)}^{2}}}\] =$-2.18\times {{10}^{-18}}J$
The energy of stationary state for n=2
${{E}_{2}}$=$=-2.18\times {{10}^{-18}}J\dfrac{1}{{{(2)}^{2}}}=-0.545\times {{10}^{-18}}J$

The energy gap between two orbits is given by equation
$\Delta E={{R}_{H}}\left( \dfrac{1}{{{n}_{1}}^{2}}-\dfrac{1}{{{n}_{f}}^{2}} \right)$
$\begin{align}
  & \Delta E=-2.18\times {{10}^{-18}}J\left( \dfrac{1}{9}-\frac{1}{1} \right) \\
 & \\
\end{align}$
\[\begin{align}
  & \Delta E=-2.18\times {{10}^{-18}}J\times \dfrac{-8}{9}=1.94\times {{10}^{-18}}J \\
 & \\
\end{align}\]
So, In Bohr’s atomic model, an electron jumps from n=1 to n=3, the amount of energy absorbed is $\begin{align}
  & 1.94\times {{10}^{-18}}J \\
 & \\
\end{align}$.

Note: -when electron moves from higher stationary state to lower stationary state, energy is emitted when an electron jump from one orbit to another orbit the amount of energy released or absorbed is given by
 \[\begin{align}
  & h\nu ={{E}_{n3}}-{{E}_{n1}} \\
 & {{E}_{n}}=\dfrac{-13.6}{{{n}^{2}}}eV \\
\end{align}\]
So, the amount of energy absorbed or released in electron volts can be calculated using this formula.