**Answer the following question.**

Calculate the de-Broglie wavelength associated with the electron revolving in the first excited state of the hydrogen atom. The ground state energy of the hydrogen atom is – 13.6 eV.

#### Solution

de-Broglie wavelength `lambda = h/(mv) = h/p`, where p is momentum of electron

Kinetic energy (KE) and momentum (p) are related by, `"KE" = p^2/(2m) ("m"="mass")`

⇒ `p = sqrt(2m("KE")`

⇒ `lambda = h/sqrt(2m("KE")`

According to Bohr's model, `"Kinetic Energy of" e^- = |"Total Energy of " e^- |=|-(13.6xxZ^2)/(n^2)|e"V"`

for Hydrogen Z = 1 and first excited state implies n = 2

`"KE" = (13.6xx1^2)/(2^2) = 3.4 "eV"`

`= 3.4 xx 1.6 xx 10^-19 "J"`

`= 5.44 xx 10^-19 "J"`

putting the values in formula for wavelength we get,

`lambda = h/sqrt(2m("KE"))=(6.63xx10^-34)/sqrt(2xx9.1xx10^-31xx5.44xx10^-19)=6.66xx10^-10"m" = 6.66Å`