91影视

Before the jump n=2, the electron effectively orbits an effective atomic number of$Z_e=Z-4.5$, where Z is the number of protons.

After the jump $n=1$, the electron effectively orbits an effective atomic number of $Z_e=Z=0.5$.

For a hydrogen like atom, the energy $E_n$is given by

$E_n=-13.6\text{eV}\frac{Z_e^2}{n^2}$

Where is the principal quantum number.

In terms of the wavelength of the photon , the energy difference before and after the jump$螖E$ is given by

$螖E=\frac{hc}{位}$

Where his Planck's constant and is the speed of light in vacuum.

When$n=2$, then $Z_2=Z-4.5$and when $n=1$then$Z_1=Z-0.5$.

The expression for $\frac{1}{位}$is calculated as follows:

$$\begin{gathered} \frac{hc}{位}=13.6\left[{\left(\frac{Z_1}{n_1}\right)}^2-{\left(\frac{Z_2}{n_2}\right)}^2\right] \\ \frac{hc}{位}=13.6\left[{\left(\frac{Z-0.5}{1}\right)}^2-{\left(\frac{Z-4.5}{2}\right)}^2\right] \\ \frac{hc}{位}=\frac{13.6}{4}\left[4\left(Z^2-Z+0.25\right)-\left(Z^2-9Z+20.25\right)\right] \\ \frac{hc}{位}=3.4\left(4Z^2-4Z+1-Z^2+9Z-20.25\right) \end{gathered}$$

Simplify further as shown below.

$$\begin{gathered} \frac{hc}{位}=3.4\left(3Z^2+5Z-19.25\right) \\ \frac{1}{位}=\left(\frac{3.4}{hc}\right)\left(3Z^2+5Z-19.25\right) \\ \frac{1}{位}=\left(\frac{3.4}{1240}\right)\left(3Z^2+5Z-19.25\right) \\ \frac{1}{位}=\left(2.74脳{10}^{-3}\right)\left(3Z^2+5Z-19.25\right) \end{gathered}$$

On comparison of the calculated values with the theoretical values:

It is visible from Table (1), the theoretical value and the experimental values are reasonably close.