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soldi70 [24.7K]
3 years ago
11

What is the highest energy photon that can be absorbed by a ground-state hydrogen atom without causing ionization?

Physics
1 answer:
lakkis [162]3 years ago
3 0

Highest energy photon absorbed: 2.18\cdot 10^{-18}J

Explanation:

An atom is said to be (positively) ionised when it absorbs a photon, and as a consequence, an electron becomes energetic enough to escape the atom, leaving an excess of positive charge behind.

In order for the electron to escape, the energy of the absorbed photon must be exactly equal to the (negative) energy of the level in which the electron lies.

For an hydrogen atom, the energy levels are given by

E_n = -13.6 \frac{1}{n^2}

where this energy is measured in electronvolts, and n is the number of the energy level.

Since the energy is negative, this means that the electron which requires most energy is the one lying in the ground state (n=1). Therefore, for an electron in the ground state, the most energy that can be absorbed from the incoming photon is

E_1 = -13.6 eV

Converting into Joules, this is equal to

E_1 = 13.6 \cdot 1.6\cdot 10^{-19}=2.18\cdot 10^{-18}J

Learn more about hydrogen atom:

brainly.com/question/2757829

#LearnwithBrainly

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A CAR ACCElerates FROM REST To
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<u>666.6 kW</u>

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  • Power = Force × Velocity

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  • F = 2,000 (20/3)
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A loaded ore car has a mass of 950 kg. and rolls on rails ofnegligible friction. It starts from rest ans is pulled up a mineshaf
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(a) 10241 W

In this situation, the car is moving at constant speed: this means that its acceleration along the direction parallel to the slope is zero, and so the net force along this direction is also zero.

The equation of the forces along the parallel direction is:

F - mg sin \theta = 0

where

F is the force applied to pull the car

m = 950 kg is the mass of the car

g=9.8 m/s^2 is the acceleration of gravity

\theta=30.0^{\circ} is the angle of the incline

Solving for F,

F=mg sin \theta = (950)(9.8)(sin 30.0^{\circ})=4655 N

Now we know that the car is moving at constant velocity of

v = 2.20 m/s

So we can find the power done by the motor during the constant speed phase as

P=Fv = (4655)(2.20)=10241 W

(b) 10624 W

The maximum power is provided during the phase of acceleration, because during this phase the force applied is maximum. The acceleration of the car can be found with the equation

v=u+at

where

v = 2.20 m/s is the final velocity

a is the acceleration

u = 0 is the initial velocity

t = 12.0 s is the time

Solving for a,

a=\frac{v-u}{t}=\frac{2.20-0}{12.0}=0.183 m/s^2

So now the equation of the forces along the direction parallel to the incline is

F - mg sin \theta = ma

And solving for F, we find the maximum force applied by the motor:

F=ma+mgsin \theta =(950)(0.183)+(950)(9.8)(sin 30^{\circ})=4829 N

The maximum power will be applied when the velocity is maximum, v = 2.20 m/s, and so it is:

P=Fv=(4829)(2.20)=10624 W

(c) 5.82\cdot 10^6 J

Due to the law of conservation of energy, the total energy transferred out of the motor by work must be equal to the gravitational potential energy gained by the car.

The change in potential energy of the car is:

\Delta U = mg \Delta h

where

m = 950 kg is the mass

g=9.8 m/s^2 is the acceleration of gravity

\Delta h is the change in height, which is

\Delta h = L sin 30^{\circ}

where L = 1250 m is the total distance covered.

Substituting, we find the energy transferred:

\Delta U = mg L sin \theta = (950)(9.8)(1250)(sin 30^{\circ})=5.82\cdot 10^6 J

8 0
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