Question

Calculate the de-Broglie wavelength associated with the electron
revolving in the first excited state of hydrogen atom. The ground state
energy of the hydrogen atom is -13.6 eV.​

Answer 1

Answer:

see the attachment given

Answer 2

The de-Broglie wavelength associated with the electron  revolving in the first excited state of hydrogen atom is $\lambda=6.65\times 10^{-10}\ m\ .$

Given :

The ground state  energy of the hydrogen atom is -13.6 eV.

We know , Energy of any state is given by :

$E=-13.6\dfrac{z^2}{n^2}$

For first excited state of hydrogen atom . putting z = 1 and n = 2 .

$E=-3.4\ .$

We know , K.E is equal to , $K.E=-E=3.4\ .$

We know, wavelength is given by :

$\lambda=\dfrac{h}{\sqrt{2mK.E}}$

Putting value of  $h=6.626\times 10^{-34}\ m^2\ kg/ s$ and mass of electron $m_e=9.1\times 10^{-31}$.

We get ,

$\lambda=\dfrac{6.626\times 10^{-34}}{\sqrt{2\times9.1\times 10^{-31}\times 3.4\times 1.6\times 10^{-19}}}\\\\\\\lambda=6.65\times 10^{-10}\ m\ .$

Hence , this is the required solution .

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Atomic structure

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