Answer:
the final kinetic energy is 0.9eV
Explanation:
To find the kinetic energy of the electron just after the collision with hydrogen atoms you take into account that the energy of the electron in the hydrogen atoms are given by the expression:
you can assume that the shot electron excites the electron of the hydrogen atom to the first excited state, that is
![E_{n_2-n_1}=-13.6eV[\frac{1}{n_2^2}-\frac{1}{n_1^2}]\\\\E_{2-1}=-13.6eV[\frac{1}{2^2}-\frac{1}{1}]=-10.2eV](https://tex.z-dn.net/?f=E_%7Bn_2-n_1%7D%3D-13.6eV%5B%5Cfrac%7B1%7D%7Bn_2%5E2%7D-%5Cfrac%7B1%7D%7Bn_1%5E2%7D%5D%5C%5C%5C%5CE_%7B2-1%7D%3D-13.6eV%5B%5Cfrac%7B1%7D%7B2%5E2%7D-%5Cfrac%7B1%7D%7B1%7D%5D%3D-10.2eV)
-10.2eV is the energy that the shot electron losses in the excitation of the electron of the hydrogen atom. Hence, the final kinetic energy of the shot electron after it has given -10.2eV of its energy is:
Answer:
At 3.86K
Explanation:
The following data are obtained from a straight line graph of C/T plotted against T2, where C is the measured heat capacity and T is the temperature:
gradient = 0.0469 mJ mol−1 K−4 vertical intercept = 0.7 mJ mol−1 K−2
Since the graph of C/T against T2 is a straight line, the are related by the straight line equation: C /T =γ+AT². Multiplying by T, we get C =γT +AT³ The electronic contribution is linear in T, so it would be given by the first term: Ce =γT. The lattice (phonon) contribution is proportional to T³, so it would be the second term: Cph =AT³. When they become equal, we can solve these 2 equations for T. This gives: T = √γ A .
We can find γ and A from the graph. Returning to the straight line equation C /T =γ+AT². we can see that γ would be the vertical intercept, and A would be the gradient. These 2 values are given. Substituting, we f ind: T =
√0.7/ 0.0469 = 3.86K.
