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Q. The de-Broglie wavelength associated with electron of hydrogen atom in this ground state is

KCETKCET 2020Atoms

Solution:

De-Broglie wavelength is given by
$\lambda=12.27 / \sqrt{ E }_{0}$,
where $E _{0}$ is the ground state energy of the hydrogen atom
whose value is $13.6\, V$
$\therefore \lambda=12.27 /(\sqrt{1} 3.6)=3.33 \,\mathring{A}$