Ask question

Ask question

All categories

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

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

You might be interested in

Two blocks of masses m and M are connected by a string and pass over a frictionless pulley. Mass m hangs vertically, and mass M

horsena [70]

Answer:

$sin\theta - \mu_k cos\theta = \frac{m}{M}$

$sin\theta - \mu_k cos\theta = 1$

Explanation:

Force of friction on M mass so that it will move down the inclined plane is given as

$F_f = \mu Mgcos\theta$

now if it is moving down the inclined plane at constant speed

so we will have

$Mgsin\theta - T - \mu mgcos\theta = 0$

on other side the mass "m" will go up at constant speed

so we have

$T - mg = 0$

so we have

$Mgsin\theta = \mu Mgcos\theta + mg$

so we have

$sin\theta - \mu_k cos\theta = \frac{m}{M}$

for special case when m = M

then we have

$sin\theta - \mu_k cos\theta = 1$

5 0

3 years ago

A 2430 pound roller coaster starts from rest and is launched such that it crests a 105 ft high hill with a speed of 59 mph. The

Rudik [331]

Answer:

Explanation:

Given

Weight of roller coaster is $W=2430\ pound$

mass of roller coaster $m=\frac{W}{g}=\frac{2430}{32.2}=75.45$

Distance traveled by roller coaster $d=396\ ft$

drag force $f_d=85\ pounds$

velocity at top $v=59 mph\approx 86.53\ ft/s$

Suppose E is the initial energy

Conserving Energy at bottom and top

$E=\frac{1}{2}mv^2+mgh+f_d\cdot d$

$E=0.5\times 75.45\times 86.53^2+2430\times 105+85\times 396$

$E=2.9\times 10^5\ foot-pound$

5 0

3 years ago

A lamp hangs from the ceiling at a height of 2.9 m. If the lamp breaks and falls to the floor, what is its impact speed

ser-zykov [4K]

To solve this there is this website that I found that helps
I am in middle school so I have no idea how to solve this
but
this website may help considering u are in high school and u
(hopefully mind u)
know how to solve this
so to get there u google
"whats impact speed"
and click on the first thing there the website is ehow

8 0

3 years ago

Projectiles that strike objects are good examples of inelastic collisions. A 0.1 kg nail driven by a gas powered nail driver col

Ratling [72]

In an inelastic collision, only momentum is conserved, while energy is not conserved.

1) Velocity of the nail and the block after the collision
This can be found by using the total momentum after the collisions:
$p_f=(m+M)v_f=4.8 kg m/s$
where
m=0.1 kg is the mass of the nail
M=10 kg is the mass of the block of wood
Rearranging the formula, we find $v_f$ , the velocity of the nail and the block after the collision:
$v_f= \frac{p_f}{m+M}= \frac{4.8 kg m/s}{0.1 kg+10 kg}= 0.48 m/s$

2) The velocity of the nail before the collision can be found by using the conservation of momentum. In fact, the total momentum before the collision is given only by the nail (since the block is at rest), and it must be equal to the total momentum after the collision:
$p_i = mv_i = p_f$
Rearranging the formula, we can find $v_i$ , the velocity of the nail before the collision:
$v_i = \frac{p_f}{m}= \frac{4.8 kg m/s}{0.1 kg}=48 m/s$

6 0

3 years ago

Read 2 more answers

what times are the moon phases visible? My science teacher said that the phases of the moon are only visible at sertant times of

klio [65]

Your teacher is right. The moon can be seen early in the morning sometimes and late at night. Different phases are only visible on certain days as one day might be full quarter, the next full moon, the next first quarter, etc.

6 0

3 years ago