How does an object behave in differents liquids?
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
Answer:
diameter of largest orbit is 0.60 m
Explanation:
given data
isotopes accelerates KE = 6.5 MeV
magnetic field B = 1.2 T
to find out
diameter
solution
first we find velocity from kinetic energy equation
KE = 1/2 × m×v² ........1
6.5 × 1.6 × = 1/2 × 1.672 ×
×v²
v = 3.5 × m/s
so
radius will be
radius = ........2
radius =
radius = 0.30
so diameter = 2 × 0.30
so diameter of largest orbit is 0.60 m
Answer:
The atomic mass unit is 1/12 of an atom of carbon 12, and is a very small amount to represent in kilograms:
is atomic mass unit.
This is why the benefits of the atomic mass unit is that it makes the representation of atomic masses easier in terms of the simplicity of the numbers that are used to represent the masses. Also using the atomic mass unit it is easier to compare the masses of different atoms, These numbers would be very small and would require negative powers of 10 to represent them, so it is more convenient to use the atomic mass unit.