# An electron in a hydrogen atom in its ground state absorbs twice of its ionization energy what is the wavelength of the emitted electron?

Jul 2, 2017

3.33 * 10^(-10 $\text{m}$

#### Explanation:

As you know, the ionization energy is the energy needed to remove $1$ mole of the outermost electrons from $1$ mole of gaseous atoms in the ground state.

${\text{X"_ ((g)) + "I.E." -> "X"_ ((g))^(+) + "e}}^{-}$

Hydrogen has an ionization energy of ${\text{1312 kJ mol}}^{- 1}$, which means that in order to remove $1$ mole of valence electrons from $1$ mole of gaseous hydrogen atoms in the ground state, you need to provide $\text{1312 kJ}$ of energy.

${\text{H"_ ((g)) + "1312 kJ" -> "H"_ ((g))^(+) + "e}}^{-}$ Now, to find the energy needed to remove the electron from a single gaseous hydrogen atom, use Avogadro's constant.

1312 "kJ"/color(red)(cancel(color(black)("mol"))) * (10^3color(white)(.)"J")/(1color(red)(cancel(color(black)("kJ")))) * (1color(red)(cancel(color(black)("mole e"^(-)))))/(6.022 * 10^(23)color(white)(.)"e"^(-)) = 2.179 * 10^(-18) $\text{J}$

So, you know that you need $2.179 \cdot {10}^{- 18}$ $\text{J}$ of energy in order to completely remove the electron from a gaseous hydrogen atom.

In your case, the electron absorbs twice as much energy, so you can say that the difference between the energy absorbed by the electron and the ionization energy will give you the kinetic energy of the emitted electron.

${K}_{\text{E" = 2 * "I.E." - "I.E.}}$

${K}_{\text{E" = "I.E.}}$

${K}_{\text{E}} = 2.179 \cdot {10}^{- 18}$ $\text{J}$

The kinetic energy of the electron is equal to

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{{K}_{\text{E}} = \frac{1}{2} \cdot m \cdot {v}^{2}}}}$

where

• $m$ is the mass of the electron
• $v$ is its velocity

Rearrange this equation to find the velocity of the electron in terms of its kinetic energy and of its mass

K_"E" = 1/2 * m * v^2 implies v = sqrt( (2 * K_"E")/m)" "color(darkorange)("(*)")

Now, the de Broglie wavelength of the electron depends on the momentum of the electron, $p$, as described by the equation

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{l a m {\mathrm{da}}_{\text{matter}} = \frac{h}{p}}}} \to$ the de Broglie wavelength

Here

• $l a m {\mathrm{da}}_{\text{matter}}$ is its de Broglie wavelength
• $h$ is Planck's constant, equal to $6.626 \cdot {10}^{- 34} \text{J s}$

On the other hand, the momentum of the electron depends on its mass and on its velocity

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{p = m \cdot v}}}$

This means that the de Broglie wavelength can be written as

$l a m {\mathrm{da}}_{\text{matter}} = \frac{h}{m \cdot v}$

Use equation $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\text{(*)}}$ to rewrite this as

lamda_"matter" = h/(m * sqrt((2 * K_"E")/m))

which is equivalent to

lamda_"matter" = h/sqrt(2 * K_"E" * m)

The mass of the electron is approximately equal to

${m}_{{\text{e}}^{-}} \approx 9.10938 \cdot {10}^{- 31}$ $\text{kg}$

You must also use the fact that

$\text{1 J} = 1$ ${\text{kg m"^2"s}}^{- 2}$

Finally, plug in your values to find

lamda_"matter" = (6.626 * 10^(-34)color(blue)(cancel(color(black)("kg"))) "m"^color(red)(cancel(color(black)(2)))color(green)(cancel(color(black)("s"^(-2)))) * color(green)(cancel(color(black)("s"))))/(sqrt(2 * 2.179 * 10^(-18)color(blue)(cancel(color(black)("kg")))color(red)(cancel(color(black)("m"^2)))color(green)(cancel(color(black)("s"^(-2)))) * 9.10938 * 10^(-31)color(blue)(cancel(color(black)("kg")))

lamda_"matter" = 6.626/sqrt(2 * 2.179 * 9.10938 * 10) * 10^(-34)/(sqrt(10^(-50)) $\text{m}$

lamda_"matter" = color(darkgreen)(ul(color(black)(3.33 * 10^(-10)color(white)(.)"m")))

I'll leave the answer rounded to three sig figs.