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

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Solution

Kinetic energy of an electron in an orbit in Bohr's model for Hydrogen atom is given by:
$E = \frac{13.6}{n^{2}}$

De-broglie wavelength of an electron is given by:
$λ = \frac{h}{p}$

But, $p = \sqrt{2 E m}$
$λ = \frac{h}{\sqrt{2 m E}}$

$\Rightarrow λ \propto n$

Hence, when electron jumps from excited state to the ground state, its de-broglie wavelength decreases.

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

Kinetic energy of an electron in an orbit in Bohr's model for Hydrogen atom is given by:E=13.6n2De-broglie wavelength of an electron is given by:λ=hpBut, p=√2Emλ=h√2mE⇒λ∝nHence, when electron jumps from excited state to the ground state, its de-broglie wavelength decreases.