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NemiM [27]
3 years ago
14

When setting your schedule, it is sometimes necessary to tell people "No." Please select the best answer from the choices provid

ed. t or f​
Physics
2 answers:
mart [117]3 years ago
6 0

Answer:

True Edg2020

Explanation:

Effectus [21]3 years ago
3 0

Answer:

True

Explanation:

You might be interested in
Model a hydrogen atom as a three-dimensional potential well with Uo = 0 in the region 0 &lt; x a. 283 eV <br> b. 339 eV <br> c.
denis23 [38]

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

8 0
2 years ago
A factory emits pollutants at a rate of 25 g/s. The factory is located between two mountain ranges resulting in an effective val
Lostsunrise [7]

Answer:

1.25\ \mu\text{g/m}^3

Explanation:

v = Velocity of the breeze = 4 m/s

w = Width of the valley = 5000 m

h = Height of the valley = 1000 m

Volumetric flow rate is given by

\dot{V}=vwh\\\Rightarrow \dot{V}=4\times 5000\times 1000\\\Rightarrow \dot{V}=2\times10^{7}\ \text{m}^3/\text{s}

\dot{m} = Mass flow rate of pollutant = 25 g/s = 25\times 10^6\ \mu\text{g/s}

Concentration is given by

C=\dfrac{\dot{m}}{\dot{V}}\\\Rightarrow C=\dfrac{25\times 10^6}{2\times 10^7}\\\Rightarrow C=1.25\ \mu\text{g/m}^3

The steady state concentration of pollutants in the valley, is 1.25\ \mu\text{g/m}^3.

6 0
3 years ago
Using monochromatic light of 410 nm wavelength, a single thin slit forms a diffraction pattern, with the first minimum at an ang
VLD [36.1K]

Answer:

c. 307 nm

Explanation:

angular position of first dark fringe =  λ / d ,   λ is wavelength and  d is width of slit .

(40 x π ) / 180 = 410 / d

angular position of second  dark fringe =  2 x λ / d ,   λ is wavelength and  d is width of slit .

(60 x π ) / 180 = 2 x λ / d

Dividing these equations

60 / 40 =  2 x λ / 410

λ = 307.5 nm.

5 0
4 years ago
Early psychologists were predominately all of the following except __________. A. white B. male C. European D. wealthy
yulyashka [42]

Answer: D. wealthy

Explanation: on the e2020 test its right

7 0
3 years ago
Read 2 more answers
I need help with question #8 please!!
otez555 [7]

Answer: 3.1158

Explanation: THERE

3 0
2 years ago
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