The de-broglie wavelength of an electron moving
with K.E 13.6 eV will be.
Answers
Explanation:
What is the de Broglie wavelength of an electron in the ground state of a hydrogen atom given that its kinetic energy is 13.6ev?
3 Answers

Tusharkanta Srichandan, B.Sc with Physics as core from S.C.S College, Puri,Odisha (2019)
Answered 3 years ago
Originally Answered: What is the de Broglie wavelength of the electron in the ground state of hydrogen atoms given that it's kinetic energy is 13.6 ev?
de Broglie wavelength (λ) is given by the equation
λ = h/p
where h=Planck’s constant whose value is 6.62 x 10^(−34) joule-seconds and
p = momentum of the particle(here electron)
In terms of kinetic energy(E) momentum(p) can be written as,
p=(2mE)^1/2
where m=mass of the particle.
Hence λ becomes
λ = h(2mE)^-1/2
Given here, E = 13.6 eV = 13.6×1.6×10^-19 joule
m(mass of electron)= 9.1×10^-31 kg
Putting these values in equation (1) we get ,
λ =0.332×10^(-9) meter
=3.32×10^(-10) meter
=3.32 Å