Question

When an electron in a hydrogen atom jumps from the excited state to the ground state, how would the de-Broglie wavelength associated with the electron change? Justify your answer.

Answer 1

Here we have to apply the de Broglie formula to get the answer.
As per wave-particle duality, the De Broglie wavelength is a wavelength shown in all the items in quantum mechanics which decides the probability of finding the article at a given point.The wavelength of the de Broglie particle is not directly proportional to its momentum.

Complete step by step answer:
On account of electrons going around and around the nuclei in atoms, the de Broglie waves exist as a closed loop, with the end goal that they can exist just as standing waves, and fit uniformly around the loop. Due to this necessity, the electrons in atoms circle the centre in specific arrangements, or states, which are called stationary orbits.De Broglie contemplated that matter additionally can show wave-particle duality, much the same as light, since light has properties both of a wave (it tends to be diffracted and it has a wavelength) and of a particle (it contains energy $h\nu$). And furthermore contemplated that matter would follow a similar condition for wavelength as light in particular,
$\lambda = \dfrac{h}{P}$
Where
$\lambda$ is the wavelength
$h$ is the Planck’s constant
$P$ is the linear momentum
Also,
Momentum,$p = mv$
Kinetic energy,$K.E = \dfrac{1}{2}m{v^2}$
Hence, from both the equations we get,
$p = \sqrt {2m{E_K}}$
Ratio of the hydrogen atom's wavelength as the atom jumps from the third excited state to the ground state (n=4 to n=1)
$\dfrac{{{\lambda _1}}}{{{\lambda _2}}} = \dfrac{{\dfrac{h}{{{p_1}}}}}{{\dfrac{h}{{{p_2}}}}} \\\ \Rightarrow\dfrac{{{\lambda _1}}}{{{\lambda _2}}} = \dfrac{{{p_2}}}{{{p_1}}} \\\ \Rightarrow \dfrac{{{\lambda _1}}}{{{\lambda _2}}}= \dfrac{{\sqrt {2m{E_{K2}}} }}{{\sqrt {2m{E_{K1}}} }} \\\$
${E_{Kn}} = \dfrac{{ - 13.6{Z^2}}}{{{n^2}}}$

So, the ratio when an electron in hydrogen atom jumps from the excited state to the ground state is:$\dfrac{{{\lambda _{n = 1}}}}{{{\lambda _{n = 4}}}} = \dfrac{1}{4}$

Note: Here we have to be careful while calculating the ratio. If we somehow change the values we may get a wrong answer.