Question

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

Answer 1

The value is
`\lambda   = 1.329 *10^(-9) \  m`

Step-by-step 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  =  (2 *  \pi  r)/(4)`

substituting values

`\lambda  =  ( 2 * 3.142  *  0.846 *10^(-9))/(4)`

`\lambda   = 1.329 *10^(-9) \  m`