physic question

Question

Assume hydrogen atoms in a gas are initially in their ground state.

If free electrons with kinetic energy 12.75 eV

collide with these atoms, what photon wavelengths will be emitted by the gas?

Express your answer using four significant figures. If there is more than one answer, enter each answer in ascending order separated by a comma.





my answer is 187.5,486.2,97.23,656.3,102.6,121.5 nm

but they said it is wrong

Answer 1

I will explain how to answer this, but YOU do the calculations.  Assuming the atom absorbs all the kinetic energy of the electron,  

 

E = (-13.6 + 12.75) eV = -0.85 eV

 

n² = -13.6/E = -13.6/-0.85 = 16

 

n = 4

 

The atom has to relax from the E₄ state to the ground state. The possible ways to get there (if I am not missing any) are

 

E₄ → E₁

E₄ → E₂ → E₁

E₄ → E₃ → E₁

E₄ → E₃ → E₂ → E₁

 

The DIFFERENT emissions are:

 

E₄ → E₁

E₄ → E₂

E₄ → E₃

E₃ → E₁
E₃ → E₂
E₂ → E₁

 

That makes a total of 6 possible emissions.  You also found 6 possible emissions.  Calculate the wavelength of each one from the Rydberg formula.  I will do

 

E₄ → E₁,

 

and you can do the others.

 

1/λ = R(1/1² - 1/4²) = (1.097 x 10⁷ m^-1)(0.9375) = 10,284,375 m^-1

 

λ ≅ 97.23 nm

 

This agrees with one of your answers.  You are on the right track, if you calculated the other 5 transitions in the same way.  Assuming you got the right numbers, did you enter them in ascending order, by wavelength?