Question

One of the emission spectral lines for Be31 has a wavelength of 253.4 nm for an electronic transition that begins in the state with n 5 5. What is the principal quantum number of the lower-energy state corresponding to this emission? (Hint: The Bohr model can be applied to one-electron ions. Don’t forget the Z factor: Z 5 nuclear charge 5 atomic number.)

Answer 1

Explanation:

The given data is as follows.

   $\lambda$ = 253.4 nm = $253.4 \times 10^{-9}m$      (as 1 nm = $10^{-9}$)

            $n_{1}$ = 5,        $n_{2}$ = ?

Relation between energy and wavelength is as follows.

                    E = $\frac{hc}{\lambda}$

                       = $\frac{6.626 \times 10^{-34} Js \times 3 \times 10^{8} m/s}{253.4 \times 10^{-9}}$

                       = $0.0784 \times 10^{-17}$ J

                       = $7.84 \times 10^{-19} J$

Hence, energy released is $7.84 \times 10^{-19} J$.

Also, we known that change in energy will be as follows.

     $\Delta E = -2.178 \times (Z)^{2}[\frac{1}{n^{2}_{2}} - \frac{1}{n^{2}_{1}}$

where, Z = atomic number of the given element

 $7.84 \times 10^{-19} J = -2.178 \times (4)^{2}[\frac{1}{n^{2}_{2}} - \frac{1}{(5)^{2}}$

    $\frac{7.84 \times 10^{-19} J}{34.848} = \frac{1}{n^{2}_{2}} - \frac{1}{(5)^{2}}$

      0.02 + 0.04 = $\frac{1}{n^{2}_{1}}$

                      $n_{1}$ = $\sqrt{\frac{1}{0.06}}$

                          = 4

Thus, we can conclude that the principal quantum number of the lower-energy state corresponding to this emission is n = 4.