Question
Download Solution PDFConsider an electron in a hydrogen atom, revolving in its second excited state (having radius 4.65 Å). The de-Broglie wavelength of this electron is:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
De Broglie wavelength λ, is defined as the Planck’s constant is divided by the particle’s momentum.
Formula:
It is used to calculate the wavelength and momentum.
From question, de Broglie's assumption about angular momentum is:
We know that, the de Broglie’s wavelength is given by the formula:
Now, the equation becomes,
Where,
n = Excited state = 3 (given)
r = Radius of the atom = 4.65 Å = 4.65 × 10-10 m
Calculation:
On substituting values,
⇒ λ = 9.73 × 10-10 m
∴ λ = 9.7 Å
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