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4 years ago

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

Physics

1 answer:

lakkis [162]4 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:

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Using the formula d = rt, calculate the distance a car would travel at 80 mph for 2.5 hours.

hoa [83]

Using the formula,

$d= v \times t$

Here, d is distance, v is the velocity and t is time.

Given, $v = 80 \ mile /hour$ and $t = 2.5 \ hour$ .

Substituting these values in above formula, we get

$d = 80 \ mile /hour \times 2.5 \ hour = 200 \ miles$

Thus, the distance traveled by the car is 200 miles

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3 years ago

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3 years ago

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An electromagnetic wave has a wavelength of 1.3m what is the frequency of wave

larisa86 [58]

V = f * wavelength
as we know electromagnetic wave has speed equal to light, so

3 * 10^8 = f * 1.3
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3 years ago

What is the minimum value of the friction coefficient between the boxes that will keep them from slipping when the 100 N force i

CaHeK987 [17]

Answer:

The friction coefficient's minimum value will be "0.173".

Explanation:

The given query seems to be incomplete. Below is the attached file of the complete question.

According to the question,

(a)

The net friction force's magnitude will be:

⇒ $F_{net}=ma$

$=5\times 1.7$

$=8.5 \ N$

(b)

For m₃,

⇒ $ma=\mu m_3 g$

Or,

⇒ $\mu=\frac{a}{g}$

$=\frac{1.7}{9.8}$

$=0.173$

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3 years ago

What aspect of motion can you conclude is common among freely falling objects ?

Liula [17]

Answer:

Free-fall is defined as the movement where the only force acting on an object is the gravitational force.

By the second Newton's law, we have that:

F = m*a

Where F = Force, m = mass, a = acceleration.

We can write this as:

a = F/m

And the gravitational force can be written as:

F = (G*M/r^2)*m

Where G is the gravitational constant, M is the mass of the Earth in this case, and r is the distance between both objects (the center of the Earth and the free-falling object)

As the radius of the Earth is really big, the term inside the parentheses is almost constant in the region of interest, then we can write:

G*M/r^2 ≈ g

And the gravitational force is:

F = g*m

And by the second Newton's law we had:

a = F/m = (g*m)/m = g

a = g

Then the acceleration does not depend on the mass of the object.

Then the thing that is common among the free-falling objects is the vertical acceleration.

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3 years ago