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Grade: 11
        
Calculate the De-Broglie Wavelength of the electron in the ground state of hydrogen atom, given kinetic energy is 13.6 eV
3 months ago

Answers : (2)

Arun
22533 Points
							

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

  1. 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

3 months ago
Vikas TU
8723 Points
							
Hiii shaow
De broglie wavelength Lambda = h/p
p in terms of kinetic energy is :
p = (2mE)^1/2
E = 13.6 eV
mass of electron = 9.1 * 10^-31
Lambda = 3.32 A0
3 months ago
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