Which element could provide one atom to make an ionic bond with nitrogen?
1 answer:
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This question is incomplete, the complete question is;
Model a hydrogen atom as a three-dimensional potential well with U₀ = 0 in the region 0 < x < L, 0 < y < L and 0 < z < L, and infinite otherwise, with L = 1.0 × 10⁻¹⁰ m.
Which of the following is NOT one of the lowest three energy levels of an electron in this model?
a. 283 eV
b. 339 eV
c. 113 eV
d. 226 eV
Answer:
the lowest three energy are; 113 eV, 225 eV, and 339 eV.
Hence Option a) 283 eV is not among the three lowest energy
Explanation:
Given the data in the question;
Three dimension cube or particle in a cubic box
the energy value is given by;
=
× π²h"² / 2ml²
where h" = h/2π and h is Planck's constant ( 6.626 × 10⁻³⁴ m² kg / s )
m is mass of electron ( 9.1 × 10⁻³¹ kg )
l is length of side of box ( 1.0 × 10⁻¹⁰ m )
for ground level ( )
so
× π²h"² / 2ml²
since h" = h/2π
× π²h² / (2π)²2ml²
so we substitute
= ( 1² + 1² + 1² ) × [ π²( 6.626 × 10⁻³⁴ )² ] / [ (2π)² × 2 × 9.1 × 10⁻³¹ kg × ( 1.0 × 10⁻¹⁰)² ]
= 3 × [ (4.333188779 × 10⁻⁶⁶) / ( 7.185072 × 10⁻⁴⁹ ) ]
= 3 × [ 6.03082165 × 10⁻¹⁸ ]
Now, we know that electric charge = 1.602 x 10⁻¹⁹
so
= 3 × [ (6.03082165 × 10⁻¹⁸) / (1.602 x 10⁻¹⁹) ]
= 3 × [ 37.645578 ]
= 112.9 ≈ 113 eV
=
× π²h² / (2π)²2ml²
we substitute
= ( 1² + 1² + 2² ) × [ 37.645578 ]
= 6 × [ 37.645578 ]
= 225.87 ≈ 226 eV
=
× π²h² / (2π)²2ml²
we substitute
= ( 2² + 2² + 1² ) × [ 37.645578 ]
= 9 × [ 37.645578 ]
= 338.8 ≈ 339 eV
Therefore, the lowest three energy are; 113 eV, 225 eV, and 339 eV.
Hence Option a) 283 eV is not among the three lowest energy
To solve the problem we will apply the concepts related to the Intensity as a function of the power and the area, as well as the electric field as a function of the current, the speed of light and the permeability in free space, as shown below.
The intensity of the wave at the receiver is
The amplitude of electric field at the receiver is
The amplitude of induced emf by this signal between the ends of the receiving antenna is
Here,
I = Current
= Permeability at free space
c = Light speed
d = Distance
Replacing,
Thus, the amplitude of induced emf by this signal between the ends of the receiving antenna is 0.0543V