physics

Question

In an hydrogen atom, an electron falls from the n = 4 level to the n = 2 level.

 

 

Use the energy level diagram for hydrogen above to help you answer the following questions:

a) During this transition, is a photon emitted or absorbed?

b) What is the change in energy of the electron, in eV's?

c) The wavelength of the emitted or absorbed photon is _____ x 10^-7 m. 

 

2 -a)What is the shortest wavelength photon that is emitted in the hydrogen atom?

b)What transition emits this photon?

Hint: If the wavelength is small, then the energy is large. Looking at the energy level diagram will also help you.

explain your answer.

Put your numerical answer to 3 significant digits and proper units.

Answer 1

1a)

During this transition, photon jumps from the higher energy band (n=4) to the lower energy band (n=2).

Energy is emitted when the electron moves to an orbit of lower energy state.

So photon emitted energy.

Answer 2

1b)

Initially electron at n = 4 level.

So initial energy E_i = - 0.85

At the end electron is at n = 2 level.

So final energy E_f = - 3.4

The change in energy of the electron E = E_f - E_i

E = – 3.4 – ( –0.85 ) = – 3.4 + 0.85 = – 2.55 eV.

The change in energy of the electron is  – 2.55 eV.

 

Answer 3

1c)

E = 2.55 eV  ( From problem 1b )

1 eV = 1.60217657 × 10^-19 J

E = 2.55 eV = 2.55×1.60217657 × 10^-19 = 4.08555 × 10^-19  J

Formula : E = hc/ λ

λ = hc/E

Where E = energy.

          c = Velocity of light = 3 × 10⁸ m/s

          h = Plank’s constant = 6.63 × 10^-34 J.s

Substitute :  c = 3 × 10⁸  , h =  6.63 × 10^-34 and E = 4.08555 × 10^-19

λ = (6.63 × 10^-34)( 3 × 10⁸)/( 4.08555 × 10^-19  )

λ = (6.63 × 3/4.08555)(10^-34+8+19)

λ = 4.8684 × 10^-7 m.

The wavelength of the emitted photon is 4.8684 × 10^-7 m.

Answer 4

2a)

There are only four state transitions that will result in a Lyman photon being emitted :

Those states are

1) n = ∞  →   n = 1

2) n = 4   →   n = 1

3) n = 3   →   n = 1

4) n = 2   →   n = 1

The wavelength of the photons is given by

R is constant. So R value will not impact on the value of λ.

The shortest λ will be given for the largest value of (1/n₁² - 1/n₂²).

If n₁ is small then (1/n₁²) is large. So n₁ must be as small as possible.

n₁ = 1   ( by considering  Lymen series )

If n₂ is high then (1/n₂²) is small. So n₂ must be as large as possible.

n₂ = ∞

Substitute Rydberg Constant value : R = 1.097373×10⁷ m^-1

Infinity value of n₂ corresponds to the ionized state of the hydrogen atom.

The shortest wavelength photon that is emitted in the hydrogen atom is 0.911×10^-7  m.

Answer 5

2b)

The shortest wavelength photon that is emitted in the hydrogen atom is at

the transition from the energy band n = ∞  to the  energy band  n = 1.