Question

What is the ratio of the Rydberg constant for helium to hydrogen atom?

• A. $2/1$
• B. $4/1$
• C. $8/1$
• D. $16/1$

Answer 1

The correct option is C $8/1$

We can use formula $R=\frac{2{\pi }^2\mu Z^2{e}^4}{(4\pi {ϵ}_0{)}^2{h}^2}$ Where$\mu$= reduce mass of atom, Z = Atomic no.

$\mu =$ $\frac{m_{nucleus}\times m_e}{m_e+m_{nucleus}}$

So, $\frac{R_{He}}{R_H}=\frac{\frac{4\times 2m_e}{m_e+4}\times 2^2}{\frac{1\times m_e}{1+m_e}\times 1^2}$ = $\frac{8}{1}$ (As me<<<1, thus $m_e+4\approx 4$, and $m_e+1\approx 1$)