Question

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

Answer 1

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

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

    $\lambda = 1.329 *10^{-9} \ m$