The charge on the electron and proton is reduced to half. Let the present value of the Rydberg const

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The charge on the electron and proton is reduced to half. Let the present value of the Rydberg constant be R. What will be the new value of the Rydberg constant?

Accepted Answer

$R \propto e^{4}$$	herefore\frac{R_{2}}{R_{1}} = \left( \frac{e_{2}}{e_{1}} ight)^{4} \Rightarrow \frac{R_{2}}{R} = \left( \frac{e/2}{e} ight)^{4} \Rightarrow R_{2} = \frac{R}{16}$

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