91影视

$$\begin{gathered} =鈭�\frac{1}{n^2}\left(\frac{m_e{e}^4}{8{系}_0^2{h}^2}\right)鈫�the\mathrm{unit}\mathrm{is}\mathrm{by}\mathrm{joule} \\ =鈭�\frac{1}{n^2}\left(\frac{m_e{e}^3}{8{系}_0{h}^2}\right)鈫�unitby\mathrm{eV} \\ =鈭�\frac{1}{n^2}\left(\frac{\left(9.1094鈭�{10}^{鈭�31}\mathrm{kg}\right){\left(1.602鈭�{10}^{鈭�19}\mathrm{c}\right)}^3}{8{\left(8.8543鈭�{10}^{鈭�12}\mathrm{N}鈰�{\mathrm{m}}^2/{\mathrm{c}}^2\right)}^2{\left(6.626鈭�{10}^{鈭�34}\mathrm{J}.\mathrm{s}\right)}^2}\right) \\ {E}_n=鈭�\frac{13.6057}{n^2}\mathrm{eV} \\ {\mathrm{E}}_n=鈭�\frac{m{e}^4}{8{系}_0^2{n}^2{h}^2} \end{gathered}$$

Planck鈥檚 Constant;

$E=hf=\frac{hc}{位}$

$$\begin{gathered} E_{纬}=\frac{hc}{位} \\ =\frac{\left(4.136脳{10}^{鈭�15}\mathrm{eV}鈰�\mathrm{s}\right)\left(3.0脳{10}^8\mathrm{m}/\mathrm{s}\right)}{85.5脳{10}^{鈭�9}\mathrm{m}}=14.5\mathrm{eV} \end{gathered}$$

Some of this energy will be converted into kinetic energy, while the rest will be used to free an electron from the atom.

(From n=1 to n=$鈭�$), the energy required to liberate an electron is:

$$\begin{gathered} \mathrm{螖}E=E_{\mathrm{鈭�}}鈭�E_1 \\ =0鈭�(鈭�13.6\mathrm{eV}) \\ =13.6\mathrm{eV} \end{gathered}$$

As a result, the photoelectron's kinetic energy is

$$\begin{gathered} K=14.5eV-13.6eV \\ =0.90eV \end{gathered}$$

The maximum kinetic energy in electron volts of these photoelectrons is 0.90eV.

$$\begin{gathered} \mathrm{螖}E=E_2鈭�E鈭�1 \\ =\frac{鈭�13.6\mathrm{eV}}{2^2}鈭�\frac{鈭�13.6\mathrm{eV}}{1^2} \\ =10.2\mathrm{eV} \end{gathered}$$

Hence, when an atom is in the excited state n=2 and is hit by a 1.45eV photon, it will gain 10.2eV more kinetic energy than the maximum kinetic energy in part (a)