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julia-pushkina [17]
2 years ago
9

Model a hydrogen atom as a three-dimensional potential well with Uo = 0 in the region 0 < x a. 283 eV b. 339 eV c.

113 eV
d. 226 eV
Physics
1 answer:
denis23 [38]2 years ago
8 0

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;

E_{nx,ny,nz = ( n_x^2 + n_y^2 + n_z^2 ) × 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 ( n_x = n_y = n_z = 1 )

so

( n_x^2 + n_y^2 + n_z^2 ) ×  π²h"² / 2ml²

since h" = h/2π

( n_x^2 + n_y^2 + n_z^2 ) × π²h² / (2π)²2ml²

so we substitute

E_{111 = ( 1² + 1² + 1² ) × [ π²( 6.626 × 10⁻³⁴ )² ] / [ (2π)² × 2 × 9.1 × 10⁻³¹ kg × ( 1.0 × 10⁻¹⁰)² ]

E_{111 = 3 × [ (4.333188779 × 10⁻⁶⁶) / ( 7.185072 × 10⁻⁴⁹ ) ]    

E_{111 = 3 × [ 6.03082165 × 10⁻¹⁸ ]

Now, we know that electric charge = 1.602 x 10⁻¹⁹

so

E_{111 = 3 × [ (6.03082165 × 10⁻¹⁸) / (1.602 x 10⁻¹⁹) ]

E_{111 = 3 × [ 37.645578 ]

E_{111 = 112.9 ≈ 113 eV

E_{211 = ( n_x^2 + n_y^2 + n_z^2 )  × π²h² / (2π)²2ml²

we substitute

E_{211 = ( 1² + 1² + 2² ) × [ 37.645578 ]

E_{211 = 6 × [ 37.645578 ]

E_{211 = 225.87 ≈ 226 eV

E_{221 = ( n_x^2 + n_y^2 + n_z^2 )  × π²h² / (2π)²2ml²

we substitute

E_{221 = ( 2² + 2² + 1² ) × [ 37.645578 ]

E_{211 = 9 × [ 37.645578 ]

E_{211 = 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

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1) The equilibrium constant Kc for the reaction N 2(g) + O 2(g) 2NO(g) at 1200 C is 1.00x 10^-5. Calculate the molar concentrati
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Explanation:

1) N₂ + O₂ → 2 NO

Kc = [NO]² / ([N₂] [O₂])

Set up an ICE table:

\left[\begin{array}{cccc}&Initial&Change&Equilibrium\\N_{2}&0.114&-x&0.114-x\\O_{2}&0.114&-x&0.114-x\\NO&0&+2x&2x\end{array}\right]

Plug into the equilibrium equation and solve for x.

1.00×10⁻⁵ = (2x)² / ((0.114 − x) (0.114 − x))

1.00×10⁻⁵ = (2x)² / (0.114 − x)²

√(1.00×10⁻⁵) = 2x / (0.114 − x)

0.00316 = 2x / (0.114 − x)

0.00361 − 0.00316x = 2x

0.00361 = 2.00316x

x = 0.00018

The volume is 1.00 L, so the concentrations at equilibrium are:

[N₂] = 0.114 − x = 0.11382

[O₂] = 0.114 − x = 0.11382

[NO] = 2x = 0.00036

2(a) Cl₂ → 2 Cl

Kc = [Cl]² / [Cl₂]

\left[\begin{array}{cccc}&Initial&Change&Equilibrium\\Cl_{2}&2.0&-x&2.0-x\\Cl&0&+2x&2x\end{array}\right]

1.2×10⁻⁷ = (2x)² / (2 − x)

1.2×10⁻⁷ (2 − x) = 4x²

2.4×10⁻⁷ − 1.2×10⁻⁷ x = 4x²

2.4×10⁻⁷ ≈ 4x²

x² ≈ 6×10⁻⁸

x ≈ 0.000245

2x ≈ 0.00049

2(b) F₂ → 2 F

Kc = [F]² / [F₂]

\left[\begin{array}{cccc}&Initial&Change&Equilibrium\\F_{2}&2.0&-x&2.0-x\\F&0&+2x&2x\end{array}\right]

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1.2×10⁻⁴ (2 − x) = 4x²

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2.4×10⁻⁴ ≈ 4x²

x² ≈ 6×10⁻⁵

x ≈ 0.00775

2x ≈ 0.0155

F₂ dissociates more, so Cl₂ is more stable at 1000 K.

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