A 10.0kg water balloon is dropped from a height of 12.0m. Calculate the speed of the balloon just before it hits the ground
1 answer:
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Answer:
1.98V
Explanation:
V =IR
V =Voltage
I =Current
R = Resistance
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