# Question #b44a1

##### 1 Answer

#### Explanation:

The key here is the kinetic energy of the electron, which can be expressed as

#color(blue)(ul(color(black)(K_E = 1/2 * m* v^2)))#

Here

#m# is the mass of the electron#v# is its velocity

The **de Broglie wavelength** depends on the *momentum* of the electron, as given by the equation

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

Here

#p# is the momentum of the electron#lamda# is its de Broglie wavelength#h# is Planck's constant, equal to#6.626 * 10^(-34)"kg m"^2"s"^(-1)#

The momentum of the electron can be expressed using the equation

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

Here

#m# is the mass of the electron#v# is its velocity

This means that the equation for the de Broglie wavelength can be rewritten as

#lamda = h/(m * v)#

Use the kinetic energy of the electron to

#K_E = 1/2 * m * v^2 implies v = sqrt( (2 * K_E)/m)#

Plug this into the above equation to find

#lamda = h/(m * sqrt( (2 * K_e)/m)) = h/sqrt( 2 * m * K_E)#

The mass of the electron is listed as

#m_"e" ~~ 9.1094 * 10^(-31)"kg"#

This means that the de Broglie wavelength of this electron is equal to

#lamda = (6.626 * 10^(-34) "kg m"^2"s"^(-1))/(sqrt(2 * 9.1094 * 10^(-31)"kg" * 2.275 * 10^(-25)"kg m"^2"s"^(-2)))#

#lamda = (6.626 * 10^(-34) color(red)(cancel(color(black)("kg"))) "m"^color(red)(cancel(color(black)(2))) color(red)(cancel(color(black)("s"^(-1)))))/(sqrt(2 * 9.1094 * 10^(-31) * 2.275 * 10^(-25)) color(red)(cancel(color(black)("kg"))) color(red)(cancel(color(black)("m"))) color(red)(cancel(color(black)("s"^(-1)))))#

#lamda = color(darkgreen)(ul(color(black)(1.029 * 10^(-6)"m")))#

The answer is rounded to four **sig figs**.