The kinetic energy of the electron orbiting in the first excited state of hydrogen atom is 3.4 eV. Determine the de- broglie wavelength associated with it.
Mass of electron=`9.1xx10^(-31)kg`,
Planck's constant=`6.63xx10^(-34)Js`.

Explanation

Kinetic energy of electron , K= 3.4 eV
`=3.4xx1.6xx10^(-19) J`
The de Broglie wavelength associated with the electron is
`lamda=h/(mv)`
where mv = momentum of electron (product of mass and velocity)
Also, `1/2mv^2=K`
`:. lamda=h/(sqrt2mK)`
`=(6.63xx10^(-34))/(sqrt(2xx9.1xx10^(-31)xx3.4xx1.6xx10^(-19)))`
`(m=9.1xx10^(-31)kg)`
`=6.66xx10^(-10)m`