# help with physics

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

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.

asked Dec 11, 2014 in PHYSICS
reshown Dec 11, 2014

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.

1b)

Initially electron at n = 4 level.

So initial energy Ei = - 0.85

At the end electron is at n = 2 level.

So final energy Ef = - 3.4

The change in energy of the electron E = Ef - Ei

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

The change in energy of the electron is  2.55 eV.

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.

2a)

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

Those states are

1) n = ∞  →   n = 2

2) n = 5   →   n = 2

3) n = 4   →   n = 2

4) n = 3   →   n = 2

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 = 2   ( by considering  Balmer series )

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

n2 = ∞

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 4/R.