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
Answer:
The magnification is ![m = 0.3674](https://tex.z-dn.net/?f=m%20%20%3D%200.3674)
Explanation:
From the question we are told that
The power of the lens is ![P = -4.00 D(dioptre)](https://tex.z-dn.net/?f=P%20%3D%20-4.00%20D%28dioptre%29)
Generally ![1 dioptre = 1 \ meter](https://tex.z-dn.net/?f=1%20dioptre%20%3D%201%20%5C%20meter)
The object distance is
the negative sign is because the distance is measured in the opposite direction of incident light (i.e away )
Generally the focal length is mathematically represented as
=>
=> ![f = 0.25 \ m](https://tex.z-dn.net/?f=f%20%3D%200.25%20%5C%20m)
converting to cm
=> ![f = 0.25 \ m = 0.25 * 100 = 25 \ cm](https://tex.z-dn.net/?f=f%20%3D%200.25%20%5C%20m%20%3D%200.25%20%2A%20100%20%3D%2025%20%5C%20cm)
Generally from lens equation we have that
![\frac{1}{f} +\frac{1}{v} -\frac{1}{u}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bf%7D%20%2B%5Cfrac%7B1%7D%7Bv%7D%20-%5Cfrac%7B1%7D%7Bu%7D)
=> ![\frac{1}{25} +\frac{1}{v} -\frac{1}{-43}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B25%7D%20%2B%5Cfrac%7B1%7D%7Bv%7D%20-%5Cfrac%7B1%7D%7B-43%7D)
=> ![v = -15.8 \ cm](https://tex.z-dn.net/?f=v%20%3D%20%20-15.8%20%5C%20cm)
Generally the magnification is mathematically represented as
![m = \frac{v}{u}](https://tex.z-dn.net/?f=m%20%20%3D%20%5Cfrac%7Bv%7D%7Bu%7D)
=> ![m = \frac{- 15.8}{-43}](https://tex.z-dn.net/?f=m%20%20%3D%20%5Cfrac%7B-%2015.8%7D%7B-43%7D)
=> ![m = 0.3674](https://tex.z-dn.net/?f=m%20%20%3D%200.3674)
The answer is Marie Skłodowska Curie (AKA Marie Curie). She <span>lived her life awash in ionizing radiation. She would be carrying bottles of the radium and polonium in the pocket of her coat and put them in her desk drawer.
So even after a century, her papers are still radioactive. Since the</span><span> most general isotope of radium, which is radium-226, has a half life of 1,601 years.</span>
1) 4°C : It has the highest density as shown on the graph.
2) Water expands when it freezes, making it less dense than just water.
3) The ice would sink to the bottom, then the rest of the water would freeze as well, the entire lake/river/whatever will freeze eliminating the organisms that live there.
That is not a question but not all scientific theories have stood the test of time