Answer:
a) n = 9.9 b) E₁₀ = 19.25 eV
Explanation:
Solving the Scrodinger equation for the electronegative box we get
Eₙ = (h² / 8m L²2) n²
where l is the distance L = 1.40 nm = 1.40 10⁻⁹ m and n the quantum number
In this case En = 19 eV let us reduce to the SI system
En = 19 eV (1.6 10⁻¹⁹ J / 1 eV) = 30.4 10⁻¹⁹ J
n = √ (In 8 m L² / h²)
let's calculate
n = √ (8 9.1 10⁻³¹ (1.4 10⁻⁹)² 30.4 10⁻¹⁹ / (6.63 10⁻³⁴)²
n = √ (98) n = 9.9
since n must be an integer, we approximate them to 10
b) We substitute for the calculation of energy
In = (h² / 8mL2² n²
In = (6.63 10⁻³⁴) 2 / (8 9.1 10⁻³¹ (1.4 10⁻⁹)² 10²
E₁₀ = 3.08 10⁻¹⁸ J
we reduce eV
E₁₀ = 3.08 10⁻¹⁸ j (1ev / 1.6 10⁻¹⁹J)
E₁₀ = 1.925 101 eV
E₁₀ = 19.25 eV
the result with significant figures is
E₁₀ = 19.25 eV
(a) The skater covers a distance of S=50 m in a time of t=12.1 s, so its average speed is the ratio between the distance covered and the time taken:
![v= \frac{S}{t}= \frac{50 m}{12.1 s}=4.13 m/s](https://tex.z-dn.net/?f=v%3D%20%5Cfrac%7BS%7D%7Bt%7D%3D%20%5Cfrac%7B50%20m%7D%7B12.1%20s%7D%3D4.13%20m%2Fs%20%20)
(b) The initial speed of the skater is
![v_i = 4 m/s](https://tex.z-dn.net/?f=v_i%20%3D%204%20m%2Fs)
while the final speed is
![v_f = 5.3 m/s](https://tex.z-dn.net/?f=v_f%20%3D%205.3%20m%2Fs)
and the time taken to accelerate to this velocity is t=2 s, so the acceleration of the skater is given by
![a= \frac{v_f - v_i}{t}= \frac{5.3 m/s-4.0 m/s}{2.0 s}=0.65 m/s^2](https://tex.z-dn.net/?f=a%3D%20%5Cfrac%7Bv_f%20-%20v_i%7D%7Bt%7D%3D%20%5Cfrac%7B5.3%20m%2Fs-4.0%20m%2Fs%7D%7B2.0%20s%7D%3D0.65%20m%2Fs%5E2%20%20)
(c) The initial speed of the skater is
![v_i = 13.0 m/s](https://tex.z-dn.net/?f=v_i%20%3D%2013.0%20m%2Fs)
while the final speed is
![v_f=0](https://tex.z-dn.net/?f=v_f%3D0)
since she comes to a stop. The distance covered is S=8 m, so we can use the following relationship to find the acceleration of the skater:
![2aS=v_f^2 -v_i^2](https://tex.z-dn.net/?f=2aS%3Dv_f%5E2%20-v_i%5E2)
from which we find
![a= \frac{-v_i^2}{2S}= \frac{-(13.0m/s)^2}{2 \cdot 8.0 m}=-10.6 m/s^2](https://tex.z-dn.net/?f=a%3D%20%5Cfrac%7B-v_i%5E2%7D%7B2S%7D%3D%20%5Cfrac%7B-%2813.0m%2Fs%29%5E2%7D%7B2%20%5Ccdot%208.0%20m%7D%3D-10.6%20m%2Fs%5E2%20%20)
where the negative sign means it is a deceleration.
We know, acceleration = final velocity - initial velocity / time
Here, if velocity is increasing, then,
Final velocity > initial velocity, in that case, acceleration is also increasing, as it is directly proportional to velocity
In short, Your Answer would be "Yes"
Hope this helps!