Question #2477

Evaluate the Rydberg constant R using Bohr theory and show that its value is R = 1.0974x10-1.

Expert's answer

The energy of electron in atom

E=(mV^{2})/2-(Ze^{2})/(4πε_{0} r.) (1)

The balance of forces

(Ze^{2})/(4πε_{0} r^{2} )=(mV^{2})/r (2)

From here we get

E=-(Ze^{2})/(8πε_{0} r.) (3)

Multiplying (2) by mr^3 and using L=nħ we get

(Ze^2 mr)/(4πε_0 )=(mVr)^2=n^2 ħ^2

And finally

E=-(Ze^4 m)/(32 ε_0^2 π^2 n^2 ħ^2.)

Then

hν=(Ze^{4} m)/(32 ε_0^{2} π^{2} ħ^{2} )(1/(n_{1}^{2} )-1/(n_{2}^{2} ))

but as we know from atomic physics

hν=hcR(1/(n_{1}^{2} )-1/(n_{2}^{2} ))

**R = (Ze**^{4} m)/(32ε_{0}^{2} π^{2} ħ^{3} c) = 1.0974

E=(mV

The balance of forces

(Ze

From here we get

E=-(Ze

Multiplying (2) by mr^3 and using L=nħ we get

(Ze^2 mr)/(4πε_0 )=(mVr)^2=n^2 ħ^2

And finally

E=-(Ze^4 m)/(32 ε_0^2 π^2 n^2 ħ^2.)

Then

hν=(Ze

but as we know from atomic physics

hν=hcR(1/(n

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