Question

Calculate the de-Broglie wavelength of the electron orbiting in the $n = 2$ states of hydrogen atom.

Answer 1

For all objects in quantum mechanics the De-Broglie wavelength is a wavelength at a given point of the configuration space which determines the probability density of finding the particular object. De-Broglie wavelength is inversely proportional to the momentum of that particular object.

Complete step-by-step answer :
The formula for finding the de-Broglie wavelength (in terms of the kinetic energy) is as follows,
$\lambda = \dfrac{h}{{\sqrt {2m{E_k}} }}$------- equation (1)
Where, $\lambda =$de-Broglie wavelength.
$m =$mass of the particle (here it is electron, so $m = 9.1 \times {10^{ - 31}}$ kg)
${E_k} =$ kinetic energy of the electron of the hydrogen atom.
And the symbol $h$ is representing a constant called planck's constant, and the value of the planck's constant is $h = 6.63 \times {10^{ - 34}}Js$
So, we need to calculate the kinetic energy possessed by the electron orbiting in $n = 2$ state.
We know the formula for the kinetic energy possessed by the electron orbiting in $n = 2$state
${E_k}$= $13.6\dfrac{{{Z^2}}}{{{n^2}}}$
Here we see that the $n = 2$and $Z = 1$ (atomic number of the hydrogen is 1)
$\Rightarrow {E_k} = \dfrac{{13.6}}{{{2^2}}} = 3.4eV$
$\Rightarrow {E_k} = 3.4 \times 1.6 \times {10^{ - 19}}J$
$\Rightarrow {E_k} = 5.44 \times {10^{ - 19}}J$
Now the de-Broglie wavelength,
Putting the values in the equation (1), we get,
$\lambda = \dfrac{{6.63 \times {{10}^{ - 34}}}}{{\sqrt {2 \times 9.1 \times {{10}^{ - 31}} \times 5.44 \times {{10}^{ - 19}}} }}$
$\Rightarrow \lambda = \dfrac{{6.63 \times {{10}^{ - 34}}}}{{\sqrt {2 \times 49.504 \times {{10}^{ - 50}}} }}$
$\Rightarrow \lambda = \dfrac{{6.63 \times {{10}^{ - 34}}}}{{\sqrt {99.008 \times {{10}^{ - 50}}} }}$
$\Rightarrow \lambda = \dfrac{{6.63 \times {{10}^{ - 34}}}}{{9.95 \times {{10}^{ - 25}}}}$
$\Rightarrow \lambda = 0.66 \times {10^{ - 9}}m$
$\Rightarrow \lambda = 0.66nm$
Hence the de-Broglie wavelength of the electron orbiting in the $n = 2$ states of hydrogen atom= $0.66nm$

Note : There is a de-Broglie equation which states that the matter may act like a wave, like light. (Since light or the radiation also acts like a wave). In other words, the de-Broglie equation suggests that the particles can also be acting like a wave. The de- Broglie equation is as follows,
$\lambda = \dfrac{h}{{mv}}$, where $\lambda$ is the de-Broglie wavelength, $h$ is Planck's constant, $m$ is mass of the particle and $v$ is the velocity with which the particle is moving.