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#Q.1# Calculate the frequency of #e^-# in the first Bohr orbit of H-atom?

Answer 1

Henry Antrim

You should get the Rydberg constant in reciprocal wavelength units as your answer.

Well, the energy levels of a hydrogen-like atom are given by:

#E_n = -(mu Z^2 e^4)/(8epsilon_0^2 h^2 n^2)#,

where:

So, the ground-state energy, i.e. the energy in the so-called "first Bohr orbit", is:

#ul(E_1) = -((9.104425135 xx 10^(-31) "kg")(1^2)(1.60217662 xx 10^(-19) cancel("C"))^cancel(4))/(8 cdot (8.854187817 xx 10^(-12) cancel("C"^2)cdot"s"^2"/"cancel"kg"cdot"m"^(cancel(3)^(2)))^cancel(2)(6.62607004 xx 10^(-34) cancel"kg"cdotcancel("m"^2)"/"cancel("s"))^cancel(2)(1^2))#

The remaining units are as expected, #"kg"cdot"m"^2"/s"^2#, or #"J"#:

#= ul(-2.17868577 xx 10^(-18) "J")#

I recall the energy should be #-"13.61 eV"#, so let's convert this to #"eV"# to check:

#E_1 = -2.17868577 xx 10^(-18) cancel"J" xx "1 eV"/(1.60217662 xx 10^(-19) cancel"J")#

#= -"13.60 eV"#

Close enough. If you used the approximation that #mu ~~ m_e#, then you would get the #-"13.61 eV"#. Now, the frequency is just given by

#E = hnu#,

so by noting that frequencies are always nonnegative... the electron would need an input of #2.17868577 xx 10^(-18) "J/atom"# to get ionized, and escape the atom with a frequency of:

#color(blue)(nu_1) = |E_1|/h = (2.17868577 xx 10^(-18) cancel"J")/(6.62607004 xx 10^(-34) cancel"J"cdot"s")#

#= color(blue)(3.288051226 xx 10^(15))# #color(blue)("s"^(-1))#

Or, perhaps a more useful unit is the #"cm"^(-1)#.

#color(blue)(R_H) = nu_1 / (2.99792458 xx 10^(10) "cm/s") = color(blue)("109677.58 cm"^(-1))#

which is, would you look at that, pretty much the Rydberg constant in #"cm"^(-1)#. Not a coincidence!

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Answer 2

Olivia Anton

Frequency = (Speed of light) / (Wavelength)
Wavelength = (Planck's constant) / (momentum of the electron in the first Bohr orbit)

Frequency ≈ 6.58 × 10^15 Hz

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Answer from HIX Tutor

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