$H_2^{+}$photodissociation. $(40$ points$)$ Let's use what we have learned about $H_2^{+}$to consider the photodissociation of $H_2^{+}.$ The molecule is initially in the ground $E_g$ state at $R=2\mathrm{.}5$bohr.

$($a$)$ Sketch the $E_g$ and $E_u$ energy surfaces $($as a function of $R,$ and use two$-$ended vertical arrows to label $E_{\text{b}\text{o}\text{n}\text{d}}=E_g-E_{1s},E_{\text{a}\text{n}\text{t}\text{i}}=E_u-E_{1s},E_{gu},$ based on the values at $R=2\mathrm{.}5.$ See below for more information about these values.

$($b$)$ Calculate the value of the overlap integral $S$ at $R=2\mathrm{.}5$bohr. Also calculate the coulomb integral $J,$ the exchange integral $K$ in hartree, for $R=2\mathrm{.}5.$

$($c$)$ From the above results, calculate $E_g,$ and calculate the bond energy in hartree, which is the $($minimum$)$ energy required to dissociate the molecule into $H(1s)$ and $H^{+},$ i$.$e$.,E_{\text{b}\text{o}\text{n}\text{d}}=E_g-E_{1s}.$

$($d$)$ Calculate the energy required for a photon to excite the molecule from $E_g$ to $E_u,$ the anti$-$bonding state, i$.$e$.,$ the optical transition energ, $E_{gu}.$

$($e$)$ Calculate the anti$-$bonding energy $E_{\text{a}\text{n}\text{t}\text{i}}=E_u-E_{1s},$ in hartree, at $R=2\mathrm{.}5$bohr, and convert to joules. In a photodissociation, this energy eventually goes to the kinetic energy of the escaping proton.

$($f$)$ Based on the above, what is the final velocity $d\frac{x}{d}t$ of the dissociated proton, in meters per second, assuming it starts in the anti$-$bonding state at $R=2\mathrm{.}5.$

$($g$)$ One way to crudely estimate the amount of time needed for the molecule to dissociate, is to assume the excess energy $(E_u-E_{1s})$ is converted instantly to kinetic energy, and calculate the time needed for the proton to move to several times the bond length. Using the velocity calculated above, estimate the amount of time $($in femtoseconds$)$ needed to move from $R=2\mathrm{.}5$ to $R=10.$

$($h$)$ Calculate the internuclear potential in the absence of an electron, at $R=2\mathrm{.}5,$ and compare to the effective potential energy of the antibonding state at the same distance, $)=(2\mathrm{.}5.$ What does this tell us about the amount of screening of internuclear charge in the excited $($antibonding$)$ state?

$($i$)$ Let's also consider photoionization. Let's assume we use an x$-$ray photon to remove the electron of an $H_2^{+}$molecule ion, at $R=2\mathrm{.}5$bohr, and the molecule $($now just two protons$)$ dissociates. What is the final $($relative$)$ velocity of the protons, and how long does it take to reach $R=10?$ Use the approach outlined above.