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:
a) 
b)
c) 
d) Treat the humans as though they were points or uniform-density spheres.
Explanation:
Given:
- mass of Mars,

- radius of the Mars,

- mass of human,

a)
Gravitation force exerted by the Mars on the human body:

where:
= gravitational constant


b)
The magnitude of the gravitational force exerted by the human on Mars is equal to the force by the Mars on human.


c)
When a similar person of the same mass is standing at a distance of 4 meters:


d)
The gravitational constant is a universal value and it remains constant in the Universe and does not depends on the size of the mass.
- Yes, we have to treat Mars as spherically symmetric so that its center of mass is at its geometric center.
- Yes, we also have to ignore the effect of sun, but as asked in the question we have to calculate the gravitational force only due to one body on another specific body which does not brings sun into picture of the consideration.
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Answer: The lithosphere and the asthenosphere