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

##### 1 Answer

#### Answer:

#### Explanation:

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

#"X"_ ((g)) + "I.E." -> "X"_ ((g))^(+) + "e"^(-)#

Hydrogen has an **ionization energy** of **mole** of valence electrons from **mole** of gaseous hydrogen atoms in the ground state, you need to provide

#"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)# #"J"#

So, you know that you need

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_"E" = 2 * "I.E." - "I.E."#

#K_"E" = "I.E."#

In your case, you have

#K_"E" = 2.179 * 10^(-18)# #"J"#

The kinetic energy of the electron is equal to

#color(blue)(ul(color(black)(K_"E" = 1/2 * m * v^2)))#

where

#m# is themassof the electron#v# is itsvelocity

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,

#color(blue)(ul(color(black)(lamda_ "matter" = h/p))) -># thede Broglie wavelength

Here

#lamda_ "matter"# is its de Broglie wavelength#h# isPlanck's constant, equal to#6.626 * 10^(-34)"J s"#

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

#color(blue)(ul(color(black)(p = m * v)))#

This means that the de Broglie wavelength can be written as

#lamda_"matter" = h/(m * v)#

Use equation

#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_ ("e"^(-)) ~~ 9.10938 * 10^(-31)# #"kg"#

You must also use the fact that

#"1 J" = 1# #"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))# #"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**.