Question

a few energy levels of the mercury atom.

a. Make a table showing all the allowed transitions in the emission spectrum. For each transition, indicate the photon wavelength, in $nm$.

b. What minimum speed must an electron have to excite the $492nm$wavelength blue emission line in the $Hg$spectrum?

Short answer

$1.8\times {10}^{4}m$ is the final answer.

Step-by-step solution

Part (a) Step 1:

To find the allowed transitions, we need to use the selection rule$∆l=\pm 1$

Using this rule we can have the following transitions:

$$\begin{gathered} 6p\to 6s,7s\to 6p,6d\to 6p,7p\to 7s \\ 8s\to 6p,8s\to 7p,8p\to 6s,8p\to 7s \\ 8s\to 8s \end{gathered}$$

Wavelength of photon:

Wavelength of the photon emitted in each transition

$\lambda =\frac{1240eVnm}{∆EeV}$

So we can have the following table:

Transition	$E_2$	$E_1$	$\lambda =\frac{1240eVnm}{(E_2-E_1)eV}$
$6p\to 6s$	$6.7eV$	$0eV$	$185nm$
$7s\to 6p$	$7.92eV$	$6.7eV$	$1015nm$
$6d\to 6p$	$8.84eV$	$6.7eV$	$580nm$
$7p\to 6s$	$8.84eV$	localid="1648630398367" $0eV$	$140nm$
$7p\to 7s$	$8.84eV$	$7.92eV$	$1348nm$
$8s\to 6p$	$9.22eV$	$6.7eV$	$492nm$
$8p\to 6d$	$9.53eV$	$8.84eV$	$1797nm$
$8p\to 6s$	$9.53eV$	$0eV$	$130nm$
$8p\to 7s$	$9.53eV$	$7.92eV$	$770nm$
$8p\to 8s$	$9.53eV$	$9.22eV$	$3999nm$

Part (b) Step 2:

A photon of wavelength $492nm$is emitted when an electron in an $Hg$atom jumps from the $8s$state to the $6p$state.

The kinetic energy of the electron must be equal to the energy of the $8s$state

$\frac{1}{2}m{v}^2=9.22eV$

$m$is the mass of an electron$=9.1\times {10}^{31}kg$

Final equation step:

$$\begin{gathered} v=\sqrt{\frac{\left(2\right)(9.22eV)(1.6\times {10}^{-19}j/eV)}{(9.1\times {10}^{-31}kg)}} \\ =1.8\times {10}^{6}m/s \end{gathered}$$