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julia-pushkina [17]
3 years ago
7

What mass of a material with density rho is required to make a hollow spherical shell having inner radius r1 and outer radius r2

? (Use any variable or symbol stated above as necessary.)
Physics
1 answer:
satela [25.4K]3 years ago
8 0

Answer:

\frac{4}{3}\rho\pi(r_2^3 - r_1^3)

Explanation:

The volume of the shell is the difference between the volume of the outer sphere and the volume of the inner sphere

V = V_o - V_i = \frac{4}{3}\pi r_2^3 - \frac{4}{3}\pi r_1^3

V = \frac{4}{3}\pi(r_2^3 - r_1^3)

The mass of the shell is the product of the volume and its density

m = V\rho = \frac{4}{3}\rho\pi(r_2^3 - r_1^3)

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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

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The wavelength of violet light is about 425 nm (1 nanometer = 1 × 10−9 m). what are the frequency and period of the light waves?
BigorU [14]

1. Frequency: 7.06\cdot 10^{14} Hz

The frequency of a light wave is given by:

f=\frac{c}{\lambda}

where

c=3\cdot 10^{-8} m/s is the speed of light

\lambda is the wavelength of the wave

In this problem, we have light with wavelength

\lambda=425 nm=425\cdot 10^{-9} m

Substituting into the equation, we find the frequency:

f=\frac{c}{\lambda}=\frac{3\cdot 10^{-8} m/s}{425\cdot 10^{-9} m}=7.06\cdot 10^{14} Hz


2. Period: 1.42 \cdot 10^{-15}s

The period of a wave is equal to the reciprocal of the frequency:

T=\frac{1}{f}

The frequency of this light wave is 7.06\cdot 10^{14} Hz (found in the previous exercise), so the period is:

T=\frac{1}{f}=\frac{1}{7.06\cdot 10^{14} Hz}=1.42\cdot 10^{-15} s


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3 years ago
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Lava cools so quickly that ions do not have time to arrange themselves into crystals will form igneous rocks with a glassy texture. Lava is the substance that flows from the volcano's. 
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