In the emission spectrum of an atom, an emission line in UV region is observed for a transition from third orbit to first orbit with the wavelength of 64.1 Angstrom. What will be the atomic number of the atom?

I tried using the formula to calculate change in energy $$\Delta E = R_h \left(\frac{1}{n_i}+\frac{1}{n_f}\right)$$ where $\Delta E$ is change in energy, $R_h$ is the Rydberg Constant $\left(2.18 \cdot 10^{-18}\right)$, $n_i$ is the initial orbit (i.e 3rd orbit) and $n_f$ is the final orbit (i.e 1st orbit).

On finding the value of this I equated it with another formula: $$E = -\frac{R_h Z^2}{n^2}$$ where $E$ is the energy, $R_h$ is the same as above, $Z$ is the atomic number and $n$ is the orbit.

I used this formula to find out the energy at 3rd orbit and then 1st orbit and took $Z^2$ common. Then I subtracted both the obtained values but failed to find out the value of $Z$ ...

  • $\begingroup$ I tried calculating change the energy using Rydberg Constant and also subtracting the energy at third orbit by the energy at first orbit, but I don't where to use the value of wavelength..... $\endgroup$
    – GanguDas
    Commented Sep 22, 2015 at 10:50
  • $\begingroup$ @GanguDas Note that the Rydberg constant actually is $R_\infty = 10\,973\,731.568\,508\ \mathrm m^{-1}$. You are using the Rydberg energy $R_\infty hc = 2.179\,872\,325 \times 10^{-18}\ \mathrm J$. $\endgroup$
    – user7951
    Commented Sep 23, 2015 at 15:39
  • $\begingroup$ @Loong yeah I need to use the second value $\endgroup$
    – GanguDas
    Commented Sep 23, 2015 at 15:55

1 Answer 1


Wikipedia (Bohr Model > Rydberg formula) tells us:

The energy of a photon emitted by a hydrogen atom is given by the difference of two hydrogen energy levels: $$E=E_i-E_f=R_\mathrm{E} \left( \frac{1}{n_{f}^2} - \frac{1}{n_{i}^2} \right)$$ where $n_f$ is the final energy level, and $n_i$ is the initial energy level.

Since the energy of a photon is $$E=\frac{hc}{\lambda}$$ the wavelength of the photon given off is given by $$\frac{1}{\lambda}=R_\infty \left( \frac{1}{n_{f}^2} - \frac{1}{n_{i}^2} \right).$$

This is known as the Rydberg formula, and the Rydberg constant $R$ is $R_\mathrm{E}/hc$, or $R_\mathrm{E}/2\pi$ in natural units.

Wikipedia also tells that it can be extended with $Z^2$ for hydrogen-like elements:

The formula above can be extended for use with any hydrogen-like chemical elements with $$\frac{1}{\lambda_{\mathrm{vac}}} = R_\infty Z^2 \left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$$ ... where $\lambda_{\mathrm{vac}}$ is the wavelength of the light emitted in vacuum, $R_\infty$ is the Rydberg constant for this element, $Z$ is the atomic number, i.e. the number of protons in the atomic nucleus of this element and, $n_1$ and $n_2$ are integers such that $n_1 < n_2$, corresponding to the principal quantum numbers of the orbitals occupied before and after.

Now you can see, that your first equation can be derived from your second equation. You also can see, that you a) forgot the squares in your first equation and b) made a sign error.

$\lambda = \mathrm{64.1 \cdot 10^{-10}\ m}$
$R_\infty = \mathrm{10973731.568508\ m^{−1}} \left(= \frac{R_E}{h\ c} = \frac{\mathrm{2.18\ 10^{-18}\ J}}{h\ c} \right)$

With all this in mind, you should be able to solve the equation to get $Z$.

$$Z^2 = \frac{n_f^2\ n_i^2}{\lambda\ R_\infty\ (n_f^2 - n_i^2)} \xrightarrow{Z>0} Z \approx 4$$


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