I know how energy of elecron and radius of orbit depends upon principal quantum number......But how it depends upon atomic number?

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    $\begingroup$ If a question is asked on Chemistry SE site, then, in contrary to sites like Quora, it is expected from the author to elaborate the topic in the question, doing at least basic research oneself, writing what he/she found and understood, and what is the stumblestone. The quick questions without explicitly expressed particular effort are not very welcome, and may be closed. $\endgroup$ – Poutnik Apr 26 '20 at 8:17

Atomic number is proportional to the atomic kernel charge. From the classical electrostatics we know that potential around a charge $Q$ is:

$$U=-\frac{Q}{4\pi \epsilon_0 r}$$

So it's obvious it the electron energy must depend on this charge.

From Schroedinger wave equation we can see the charge is involved:

$$ \psi_{n\ell m}(r,\theta,\varphi) = \sqrt {\left ( \frac{2}{n a_0} \right )^3\frac{(n-\ell-1)!}{2n[(n+\ell)!]} } e^{- r/na_0} \left(\frac{2r}{na_0}\right)^\ell L_{n-\ell-1}^{2\ell+1}\left(\frac{2r}{na_0}\right) \cdot Y_{\ell}^m(\theta, \varphi ) $$


  • $ a_0 = \frac{4 \pi \varepsilon_0 \hbar^2}{m_q q^2}$ is the Bohr radius,

  • $ L_{n-\ell-1}^{2\ell+1}(\cdots) $ are the generalized Laguerre polynomials of degree $ n - \ell -1 $.

  • $ n, \ell, m $ are the principal, azimuthal quantum and magnetic quantum numbers.

The Rydberg formula for hydrogen

....can be extended for use with any Hydrogen-like (1 electron only) chemical element with

$$\frac{1}{\lambda_{\mathrm{vac}}} = RZ^2 \left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$$

where $R$ is the Rydberg constant and $Z$ is the atomic number.

For multielectron atoms, things get complicated by shielding of kernels by inner electrons and by mutual electron repulsion.

  • $\begingroup$ Can u pls answer in a simple statement form.... $\endgroup$ – Ayush Apr 26 '20 at 8:42
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    $\begingroup$ I would turn the table: Can you ask by not a simple sentence question ? Do your figural homework first. $\endgroup$ – Poutnik Apr 26 '20 at 8:45

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