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`        Calculate the De-Broglie Wavelength of the electron in the ground state of hydrogen atom, given kinetic energy is 13.6 eV`
10 months ago

```							de Broglie wavelength (λ) is given by the equationλ = h/pwhere h=Planck’s constant whose value is 6.62 x 10^(−34) joule-seconds andp = momentum of the particle(here electron)In terms of kinetic energy(E) momentum(p) can be written as,p=(2mE)^1/2where m=mass of the particle.Hence λ becomes	λ = h(2mE)^-1/2Given here, E = 13.6 eV = 13.6×1.6×10^-19 joulem(mass of electron)= 9.1×10^-31 kgPutting these values in equation (1) we get ,λ =0.332×10^(-9) meter=3.32×10^(-10) meter=3.32 Å
```
10 months ago
```							Hiii shaowDe broglie wavelength Lambda = h/pp in terms of kinetic energy is :p = (2mE)^1/2E = 13.6 eVmass of electron = 9.1 * 10^-31Lambda = 3.32 A0
```
10 months ago
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