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:
<h3>The answer is 0.89 L</h3>
Explanation:
The new volume can be found by using the formula for Boyle's law which is
Since we are finding the new volume
From the question we have
We have the final answer as
<h3>0.89 L</h3>
Hope this helps you
Answer:
The angular speed of the Crab nebula pulsar is 190.3 rad/s.
Explanation:
Given that,
Time T= 33 ms = 0.033 s
The angular speed is equal to the 2π divided by time period.
We need to calculate the angular speed of the Crab nebula pulsar
Using formula of angular speed
Where, T = time
= angular speed
Put the value into the formula
Hence, The angular speed of the Crab nebula pulsar is 190.3 rad/s.
C. because energy can not be created or destroyed, but transformed/transferred to maintain equilibrium.
