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Levart [38]
3 years ago
11

The orbital radius of an electron in a hydrogen atom is 0.846 nm. What is its de Broglie wavelength?

Physics
1 answer:
kotykmax [81]3 years ago
6 0

Answer:

The  value  is  \lambda   = 1.329 *10^{-9} \  m

Explanation:

From the question we are told that

  The  orbital radius is  r =  0.846nm =  0.846 *10^{-9} \ m

Generally the de Broglie wavelength is mathematically represented as

      \lambda  =  \frac{2 *  \pi  r}{4}

substituting values

     \lambda  =  \frac{ 2 * 3.142  *  0.846 *10^{-9}}{4}

    \lambda   = 1.329 *10^{-9} \  m

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