Question

consider the hydrogen atom as a one - dimensional box with a length of 106 pm. calculate the wavelength of radiation emitted when its electron transitions from the n = 4 state to the n = 3 state.
λ = m

Explanation:

Step1: Recall energy - level formula for particle in a box

The energy levels of a particle in a one - dimensional box are given by $E_n=\frac{n^{2}h^{2}}{8mL^{2}}$, where $n$ is the quantum number, $h = 6.626\times10^{-34}\ J\cdot s$ is Planck's constant, $m = 9.109\times10^{-31}\ kg$ is the mass of the electron, and $L$ is the length of the box.

Step2: Calculate $E_4$ and $E_3$

$E_4=\frac{4^{2}h^{2}}{8mL^{2}}=\frac{16h^{2}}{8mL^{2}}$, and $E_3=\frac{3^{2}h^{2}}{8mL^{2}}=\frac{9h^{2}}{8mL^{2}}$.
The change in energy $\Delta E=E_4 - E_3=\frac{(16 - 9)h^{2}}{8mL^{2}}=\frac{7h^{2}}{8mL^{2}}$.
Given $L = 106\ pm=106\times10^{-12}\ m$.
Substitute $h = 6.626\times10^{-34}\ J\cdot s$, $m = 9.109\times10^{-31}\ kg$ and $L = 106\times10^{-12}\ m$ into the $\Delta E$ formula:
\[

$$\begin{align*} \Delta E&=\frac{7\times(6.626\times 10^{-34})^{2}}{8\times9.109\times10^{-31}\times(106\times10^{-12})^{2}}\\ &=\frac{7\times4.39\times10^{-67}}{8\times9.109\times10^{-31}\times1.12\times10^{-20}}\\ &=\frac{3.073\times10^{-66}}{8.096\times10^{-50}}\\ &\approx3.8\times10^{-17}\ J \end{align*}$$

\]

Step3: Use the energy - wavelength relation

The energy of a photon is $E = h
u=\frac{hc}{\lambda}$, so $\lambda=\frac{hc}{\Delta E}$.
Substitute $h = 6.626\times10^{-34}\ J\cdot s$, $c = 3\times10^{8}\ m/s$ and $\Delta E=3.8\times10^{-17}\ J$ into the formula:
\[

$$\begin{align*} \lambda&=\frac{6.626\times10^{-34}\times3\times10^{8}}{3.8\times10^{-17}}\\ &=\frac{19.878\times10^{-26}}{3.8\times10^{-17}}\\ &\approx5.23\times10^{-9}\ m \end{align*}$$

\]

Answer:

$5.23\times 10^{-9}\ m$