Question

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.

Answer 1

de Broglie wavelength, λ=h/(2mE)^1/2 where h=Plank's constant=6.626×10-34 J s m=Mass of the electron=9.1×10-31 kg E=Energy of the electron=3.4 eV=3.4×1.6×10-19
J⇒λ=6.626×10-342(9.1×10-31 )(3.4×1.6×10-19)⇒λ=0.67×10-9 m

Answer 2

Answer:

Wavelength = 665.088 pm

Explanation:

The expression for the deBroglie wavelength and kinetic energy is:

$\lambda=\frac {h}{\sqrt{2\times m\times K.E.}}$

Where,  

$\lambda$ is the deBroglie wavelength  

h is Plank's constant having value $6.626\times 10^{-34}\ Js$

m is the mass of electron having value $9.11\times 10^{-31}\ kg$

K.E. is the kinetic energy of the electron.

Given, K.E. = 3.4 eV

Energy in eV can be converted to energy in J as:

1 eV = 1.6022 × 10⁻¹⁹ J

So, K.E. = $3.4\times 1.6022\times 10^{-19}\ J=5.44748\times 10^{-19}\ J$

Applying in the equation as:

$\lambda=\frac {h}{\sqrt{2\times m\times K.E.}}$

$\lambda=\frac{6.626\times \:10^{-34}}{\sqrt{2\times \:\:9.11\times \:\:10^{-31}\times \:\:5.44748\times \:\:10^{-19}\:}}\ m$

$\lambda==\frac{10^{-34}\times \:6.626}{\sqrt{10^{-50}\times \:99.2530856}}\ m$

$\lambda=665.088\times 10^{-12}\ m$

Also, 1 m = 10¹² pm

So, Wavelength = 665.088 pm