atoms

Question

1. An electron of energy 5.3 eV collided with an atom called Tritium and was reflected with an

energy of 1.7 eV. Immediately after the atom emitted a photon. What is the frequency of the

photon?

 

2. What is the energy difference between the two energy levels in a atom that give rise to the

emission of a 440 nm photon? 

Answer 1

1).

The Energy of a electron E_e = 5.3 eV

Reflected energy E_r = 1.7 eV.

The change in energy of the electron E = E_e - E_r

E = 5.3 – 1.7 = 3.6 eV.

The change in energy of the electron is  3.6 eV

1 eV  = 1.6 x 10^-19 Joules

E = 3.6 x 1.6 x 10^-19 J

E = 5.76 x 10^-19 J 

E = hc/λ

E = h(λf)/λ = hf                           (  Since c = λf   )

f = E/h

Where,   E = Energy of Photon in Joules

             h = Plank's constant = 6.626 x10^-34.

             λ = wave length

             c = Speed of light  = 3 x10⁸ m/s.

             f = frequency of a photon in Hz

f = E/h

Substitute the values E = 5.76 x 10^-19  and h = 6.626 x10^-34 .

f = (5.76 x 10^-19)/( 6.626 x10^-34)

f = 0.8693  x 10¹⁵  Hz.

The frequency of a photon is 0.8693  x 10¹⁵  Hz.

Answer 2

2).

Formula : ΔE = hc/ λ.

Where, ΔE = energy difference.

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

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

            λ = wave length.

Substitute the values c = 3 × 10⁸m/s  , h =  6.63 × 10^-34 and  λ = 440 nm = 440 × 10^-9 m.

ΔE = [(6.63 × 10^-34)( 3 × 10⁸)] / (440 × 10^-9 )

ΔE = [(6.63 × 3)/44](10^-34+8+8)

ΔE = 0.4520 × 10^-18 J.

The energy difference between the two energy levels is 0.4520 × 10^-18 J.