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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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Answer:

a.) a = 0 ms⁻²

b.) a = 9.58 ms⁻²

c.) a = 7.67 ms⁻²

Explanation:

a.)

    Acceleration (a) is defined as the time rate of change of velocity

                       a = \frac{v_{2} - v_{1} } {t}  

Given data

 Final velocity = v₂ = 0 m/s

 Initial velocity = v ₁ = 0 m/s

  As the space shuttle remain at rest for the first 2 minutes i.e there is no change in velocity so,

                 a = 0 ms⁻²

b.)

     Given data

As the space shuttle start from rest, So initial velocity is zero

    Initial velocity = v₁ = 0 ms⁻¹

    Final velocity  = v₂ = 4600 ms⁻¹

     Time = t = 8 min = 480 s

By the definition of Acceleration (a)

             a = \frac{v_{2} - v_{1} } {t}  

             a = \frac{4600 - 0 } {480}

                     a = 9.58 ms⁻²

c.)

    Given data

As the space shuttle is at rest for first 2 min then start moving, So initial velocity is zero

    Initial velocity = v₁ = 0 ms⁻¹

    Final velocity  = v₂ = 4600 ms⁻¹

     Time = t = 10 min = 600 s

By the definition of Acceleration (a)

             a = \frac{v_{2} - v_{1} } {t}  

             a = \frac{4600 - 0 } {600}

                     a = 7.67 ms⁻²

8 0
3 years ago
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