Why does plasmon have higher erngy than phonon? In my mind plasmon is movement of electrons and phonon is movement of atoms in an lattice. movement of atoms should have a large energy because atom is larger.
 A: Fundamentally the plasmon is at higher energy because electrons weigh much less that atoms. In fact phonons are simply plasmons of the atoms.
One can see this from the classical formula for the plasma frequency.
$$\omega = \sqrt{\frac{4\pi n q^2}{m}}$$
Assuming you have an equal concentration of free electrons and atoms, $n$ will be the same for the atom and electron plasma frequencies. However,  the nuclei have masses that are usually $10^4$ to $10^5$ times larger than the electron,  which translates to a frequencies that are over 100 times smaller for phonons compared to electrons. This is consistent with experiments seeing phonons with milli-electronvolt energies and plasmons with electronvolt energies.
A: The phonon describes the vibrational motion in a solid.
The plasmon is a quasi particle, describing the plasma oscillation.
These quantized excitations have different energy levels, phonons in the meV range, and plasmon in the eV.
As you write you are wrong, because phonons energy is 3 orders of magnitude higher.
The difference is because phonons are collective excitations of atoms in lattice. Plasmons are collective excitations of nearly free electrons in plasma.
Phonons are important in conservation of energy, plasmon gives the color of gold. 
http://iopscience.iop.org/article/10.1088/0368-3281/2/1/305
A: Like it's said Plasmons describe collective oscillations of charge carriers. One can derive that the energy is given by (with $n$ the charge density, $\varepsilon$ vacuum dielectric constant): 
\begin{equation}
E_{p}=\hbar \cdot \omega_{p}=\hbar \sqrt{\frac{ne^{2}}{m\varepsilon} }
\end{equation}  
Now you can compare it to the energy of a phonon (with $n_B$ being the Bose statistic):
$$\varepsilon _{n}({\mathbf  {k}})=\hbar \cdot \omega ({\mathbf  k})\cdot \left(n_B+{\frac  {1}{2}}\right),$$
but here you have to distinguish between optical and acoustical modes (see e.g. Kittel "Solid State Physics"). You can basically excite optical phonons (which have an higher energy than the acoustical ones) with inelastic neutron scattering. 
To make a long comment short, I would suggest you can put in some numbers, but you can also argue I want to excite with same frequency (same $\omega$). Then still the Phonons have higher energy, because the Bose statistic gives you very large values for small energies. All in all the Phonons have more energy then the Plasmons. 
