Question

Wavelength of electron in ground state of hydrogen atom

Answer 1

Wavelength of electron in ground state of hydrogen atom.

1. The energy of the hydrogen atom in its ground state is -13.6eV.
2. The total energy of an electron in the hydrogen atom in ground state is -13.6eV.

Answer 2

Answer:

Explanation:

hew mate

here is your answer

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

The de broglie wavelength is given by:

λ=hp

But p(momentum)=2mK−−−−√

where K is the kinetic energy of the electron(be careful to convert the unit of KE to joule. 13.6eV=2.18*10^-18 J)

Thus the de broglie eqn can be written as:

λ=h2mK−−−−√

Now substitute the values of h,m(mass of electron) and the kinetic energy and get the answerhe velocity v of the hydrogen electron in its ground state is v = αc where α is the fine structure constant. Hence its momentum is mαc where m is the electron mass. So the de Broglie wavelength is h/mαc = 0.332 nm, same result Tusharkanta Srichandan got.

(Although I didn’t need the given KE of the electron, using the same numbers I used tacitly above for electron mass m and velocity v, its KE is ½mv² = 13.6 ev as it turns out. The reason this is consistent with the virial theorem, contrary to both Brad Barker and his upvoter Cristian Randieri, is that the PE is −27.2 ev, twice the KE to within sign, whence the total energy KE + PE = −13.6 ev.)

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