Question

What is the energy in joules required to shift the electron of the hydrogen atom from the first Bohr orbit to the fifth Bohr orbit and what is the wavelength of the light emitted when the electron returns to the ground state? The ground state electron energy is $-2.18\times {10}^{-11}ergs$.

Answer 1

Energy associated with the nth orbit of hydrogen atom is given by, $E_n=-\frac{R_H}{n^2}\{R_H=-2.18\times {10}^{-18}J\}$


Energy of emitted photon is given by

$\Delta E=hv=E_2-E_1=-R_H(\frac{1}{n_2^2}-\frac{1}{n_1^2})$

Where $E_1\text{and}{E}_2$ are energies of electrons in orbits $n1\text{and}n2$.
So, energy absorbed when the electron jumps from $1$st to $5$th orbit, will be,
$\Delta E=E_5-E_1=-R_H(\frac{1}{5^2}-\frac{1}{1^2})$

Taking $R_H=2.18\times {10}^{-18}J$, we get
$\Delta E=-2.18\times {10}^{-18}J\times (\frac{1}{25}-1)$

$\Delta E=2.18\times J\times \frac{24}{25}=2.0928\times {10}^{-18}J$

Therefore, the required to shift the elctron from $n=1\text{to}n=5\text{is}2.0928\times {10}^{-18}J$

Now,we know $\Delta E=\frac{hc}{\lambda }$

Where,$h=\text{planck's constant}=6.626\times {10}^{-34}Js$


$c=\text{velocity of light}=3\times {10}^{8}m{s}^{-1}$

$\lambda =\text{wave lenght of photon}$

Substituting the values, we get,

$2.0928\times {10}^{-18}J=\frac{6.626\times {10}^{-34}Js\times 3\times {10}^{8}m{s}^{-1}}{\lambda }$

$\Rightarrow \lambda =\frac{6.626\times {10}^{-34}Js\times 3\times {10}^{8}m{s}^{-1}}{2.0928\times {10}^{-18}J}\Rightarrow \lambda =9.498\times {10}^{-8}m\approx 950A^{\circ }$

Therefore, the wavelength of emitted radiation will be $950A^{\circ }$.
Final Answer:
Energy$=2.0928\times {10}^{-18}J$ and wave length=$950A^{\circ }$