Find the energy (in eV) of a ground state electron in a He II (He + ) atom. (b.) Find the amount of energy (in eV) needed to ion
1 answer:
Answer:
a) -54.42 eV
b) C VI (C +++++)
Explanation:
Hi!
I imagine that the formula given in class is the energy of the hydorgen like atoms, which is:
Where all are physical constants, such as vacuum permitivity ε0, planck's constant (reduced) h bar, electron's charge e, the atomic number Z, and the reduced mass μ.
The reduced mass is given by the formula:
In general the electron's mass is much smaller than the nucleous mass therefore the reduced mass is almost equal to the electron's mass.
If we replace all these constants the energy will become:
For He --> Z = 2, therefore:
And the ground state is given when n=1
a)
E = -54.42 eV
b)
This formula applies only to hydorgen like atoms, that is whit only one electron, therefore if an element were to have Z electrons, the formula will only apply to that element ionized Z-1 times.
For C Z=6
We could only calculate the quantum energy levels with the given formula for C VI (C +++++)
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Answer:
a) θ = 58.3º
b) vfh = 13.7 m/s
c) g = -9.8 m/s2
d) h = 22.2 m
e) vfb = 15.5 m/s
Explanation:
a)
- Assuming that gravity is the only influence that causes an acceleration to the water, due to it is always downward, since both directions are independent each other, in the horizontal direction, the water moves at a constant speed.
- Since the velocity vector has a magnitude of 26.0 m/s, we can find its horizontal component as follows:
- vₓ₀ = v * cos θ (1)
- where θ is the angle between the water and the horizontal axis (which we define as the x-axis, being positive to the right).
- Applying the definition of average velocity, taking the end of the hose like the origin, and making t₀ = 0, we can write the following expression:
- Replacing by the givens of xf = 41.0m, t = 3.00 s, and v=26.0 m/s, we can solve for the angle of elevation θ, as follows:
- ⇒θ = cos⁻¹ (0.526) = 58.3º (4)
b)
- At the highest point in its trajectory, just before starting to fall, the vertical component of the velocity is just zero.
- Since the horizontal component keeps constant during all the journey, we can conclude that the speed at this point is just v₀ₓ, that we can find easily from (1) replacing by the values of v and cos θ, as follows:
- vₓ₀ = v * cos θ = 26.0 m/s * 0.526 = 13.7 m/s. (5)
c)
- At any point in the trajectory, the only acceleration present is due to the action of gravity, which accepted value is -9.8 m/s2 (taking the upward direction on the vertical y-axis as positive)
d)
- Since we know the time when the water strikes the building, it will be the same for the vertical movement, so, we can use the kinematic equation for vertical displacement, as follows:
- Our only unknown remains v₀y, which can be obtained in the same way than the horizontal component:
- v₀y = v * sin θ = 26.0 m/s * 0.85 = 22.1 m/s (7)
- Replacing (7) in (6), we get:
e)
- When the water hits the building the velocity vector, has two components, the horizontal vₓ and the vertical vy.
- The horizontal component, since it keeps constant, is just v₀x:
- v₀ₓ = 13.7 m/s
- The vertical component can be found applying the definition of acceleration (g in this case), solving for the final velocity, as follows:
- Replacing by the time t (a given), g, and v₀y from (7), we can solve (9) as follows:
- Since we know the values of both components (perpendicular each other), we can find the magnitude of the velocity vector (the speed, i.e. how fast is it moving), applying the Pythagorean Theorem to v₀ₓ and v₀y, as follows: