Answer:
(a) 0.3778 eV
(b) Ratio = 0.0278
Explanation:
The Bohr's formula for the calculation of the energy of the electron in nth orbit is:
(a) The energy of the electron in n= 6 excited state is:
Ionisation energy is the amount of this energy required to remove the electron. Thus, |E| = 0.3778 eV
(b) For first orbit energy is:
Ratio = 0.0278
Answer:
Explanation:
Explain how a projectile might be modified to decrease the air resistance impacting its trajectory.
Answer: The force does not change.
Explanation:
The force between two charges q₁ and q₂ is:
F = k*(q₁*q₂)/r^2
where:
k is a constant.
r is the distance between the charges.
Now, if we increase the charge of each particle two times, then the new charges will be: 2*q₁ and 2*q₂.
If we also increase the distance between the charges two times, the new distance will be 2*r
Then the new force between them is:
F = k*(2*q₁*2*q₂)/(2*r)^2 = k*(4*q₁*q₂)/(4*r^2) = (4/4)*k*(q₁*q₂)/r^2 = k*(q₁*q₂)/r^2
This is exactly the same as we had at the beginning, then we can conclude that if we increase each of the charges two times and the distance between the charges two times, the force between the charges does not change.
Answer:........... .. .....
The answer should be d because they are constantly rotating
