What is the de Broglie wavelength of an electron in the ground state of a hydrogen atom?

1 Answer
May 20, 2017

λ = "332.5 pm"

Explanation:

The formula for the de Broglie wavelength λ is

color(blue)(bar(ul(|color(white)(a/a)λ = h/(mv)color(white)(a/a)|)))" "

where

$h =$ Planck's constant
$m =$ the mass of the electron
$v =$ the speed of the electron.

Calculate the speed of the electron

The energy $E$ of a hydrogen electron in an orbit is

$E = - {R}_{\textrm{H}} / {n}^{2}$

where

${R}_{\textrm{H}} =$ the Rydberg energy constant (2.180 × 10^"-18"color(white)(l) "J")

Since $n = 1$,

${E}_{1} = \text{-2.180 × 10"^"-18"color(white)(l) "J}$

$K E = \text{-"E_1 = 2.180 × 10^"-18"color(white)(l) "J}$

$K E = \frac{1}{2} m {v}^{2}$

v = sqrt((2KE)/m) = sqrt((2 × 2.180 × 10^"-18" color(red)(cancel(color(black)("J"))))/(9.11 × 10^"-31" color(red)(cancel(color(black)("kg")))) × (1 color(red)(cancel(color(black)("kg")))·"m"^2"s"^"-2")/(1 color(red)(cancel(color(black)("J"))))) = sqrt(4.786 × 10^12color(white)(l) "m"^2"s"^"-2") = 2.188 × 10^6color(white)(l) "m·s"^"-1"

Calculate the de Broglie wavelength

λ = h/(mv) = (6.626 × 10^"-34" color(red)(cancel(color(black)("J·s"))))/(9.109 × 10^"-31" color(red)(cancel(color(black)("kg"))) × 2.188 × 10^6 color(red)(cancel(color(black)("m·s"^"-1")))) × (1 color(red)(cancel(color(black)("kg")))·"m"^color(red)(cancel(color(black)(2)))·color(red)(cancel(color(black)("s"^"-2"))))/(1 color(red)(cancel(color(black)("J")))) = 3.325× 10^"-10"color(white)(l) "m" = "332.5 pm"