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.) the inverse of the square of the distance separating them
Explanation:
Coulombs law states that "the force of attraction between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them."
Mathematically, F = kq1q2/r²
Where q1 and q2 are the charges
r is the distance between the charges.
According to the law, the force between two charged objects is related to the inverse of the square of the distance separating them.
Answer:
its probably still trying to load ur next rank or whatever it did it to me too
Explanation:
This problem is to let you practice using Newton's second law of motion:
Force = (mass) x (acceleration)
-- The airplane's mass when it takes off (before it burns any of its load of fuel) is 320,000 kg.
-- The force available is (240,000 N/per engine) x (4 engines) = 960,000 N.
-- Now you know ' F ' and ' mass '. Use Newton's second law of motion to calculate the plane's acceleration.
Answer: The area of brick in contact with the floor is 1539 
.
Explanation:
Given: Length = 19 cm
Width = 9 cm
Height = 9 cm
As the brick is rectangular in shape. Hence, its area will be calculated as follows.
Substitute the values into above formula as follows.
Thus, we can conclude that area of brick in contact with the floor is 1539 .
