Question

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

Answer 1

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

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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:

• `mu = (m_em_p)/(m_e + m_p) = 9.104425135 xx 10^(-31) "kg"` is the reduced mass of the electron + proton.
• `m_e = 9.10938356 xx 10^(-31) "kg"` is the rest mass of the electron.
• `m_p = 1.672621898 xx 10^(-27) "kg"` is the rest mass of the proton.
• `Z` is the atomic number.
• `e` is the elementary charge, `1.60217662 xx 10^(-19) "C"`.
• `epsilon_0 = 8.854187817 xx 10^(-12) "C"^2cdot"s"^2"/kg"cdot"m"^3` is the vacuum permittivity.
• `h = 6.62607004 xx 10^(-34) "kg"cdot"m"^2"/s"` is Planck's constant.
• `n` is the principal quantum number.

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!