Question

if the total energy of an electron in a hydrogen atom in excited state is -3.4 eV, then the de Broglie wavelength of the electron is​

Answer 1

Explanation:

De Broglie's wavelength,

λ = the circumference (2πr) of orbit for n=1

That is, an electron in the first bohr orbit.

To find the circumference,

The total energy (E) of an electron in a hydrogen atom,

Energy of an electron,

E = (-) ke² / 2r  

For the ground state (n=1)

E = -13.60 eV  

E = (13.60 x 1.60^-19)J

E = 2.176^-18 J  

Substituting in the total energy formula,

2.176^-18 = ke² / 2r  

r = ke² / (2 x 2.176^-18)  

r = (9.0^9)(1.60^-19)² / (4.352^-18)

r = 5.29^-11 m  

According to De Broglie,

1λ = the circumference of orbit

So,

λ = 2πr = 2 x π x 5.29^-11 m

λ = 3.326^-10 m

The De Broglie wavelength for an electron in the first bohr orbit of a hydrogen atom is 3.326^-10 m

Answer 2

Answer:

The de Broglie wavelength of the electron is $6.653 \times 10^{-10}\ m$ .  

Explanation:

Given the total energy of an electron in hydrogen atom in excited state is -3.4 eV.

Total energy of an electron in excited state is equal to the negative of kinetic energy.

Therefore, kinetic energy $K.E=3.4eV$

                                                      $=3.4\times 1.6 \times 10^{-19} J$

The de Broglie wavelength of the electron is​ given by

                          $\lambda=\frac{h}{\sqrt{2mK.E} }$

where h is the Planck's constant,  $=6.63 \times 10^{-34} \ m^{2} kg/s$

and m is the mass of electron, $9.1 \times 10^{-31} \ kg$

                     $\therefore \lambda=\frac{6.63 \times 10^{-34} }{\sqrt{2\times 9.1 \times 10^{-31} \ \times 3.4\times 1.6 \times 10^{-19} } }$

                           $=\frac{6.63 \times 10^{-34} }{\sqrt{99.5 \times 10^{-50} } }=\frac{6.63 \times 10^{-34} }{9.95 \times 10^{-25} } }$

                           $=6.653 \times 10^{-10}\ m$

The de Broglie wavelength of the electron is $6.653 \times 10^{-10}\ m$.