crichter4098 crichter4098
  • 07-09-2020
  • Physics
contestada

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

Respuesta :

okpalawalter8
okpalawalter8 okpalawalter8
  • 08-09-2020

Answer:

The  value  is  [tex]\lambda   = 1.329 *10^{-9} \  m[/tex]

Explanation:

From the question we are told that

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

Generally the de Broglie wavelength is mathematically represented as

      [tex]\lambda  =  \frac{2 *  \pi  r}{4}[/tex]

substituting values

     [tex]\lambda  =  \frac{ 2 * 3.142  *  0.846 *10^{-9}}{4}[/tex]

    [tex]\lambda   = 1.329 *10^{-9} \  m[/tex]

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