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Question
What is the electron’s de Broglie wavelength when its momentum magnitude is \[mv_n\]?
Options
\[\lambda=\frac{h}{mv_n}\]
\[\lambda=\frac{nh}{mv_n}\]
\[\lambda=\frac{h}{mv_nr_n}\]
\[\lambda=\frac{mv_n}{h}\]
MCQ
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Solution
Substituting \[p=mv_n\] into \[\lambda=\frac{h}{p}\] gives \[\lambda=\frac{h}{mv_n}\]. This wavelength is used in the standing-wave condition for a circular orbit.
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