An electron collides with a hydrogen atom in its ground state and excites it to a state of n = 3. How much energy was given to the hydrogen atom in this inelastic collision?a)15.1 eVb)13.1 eVc)12.1 eVd)14.1 eVCorrect answer is option 'C'. Can you explain this answer?

Question

Answer

Energy at ground state E₁=−13.6 eV
Energy at n=3: E₃=−13.6/9=1.5 eV
To excite it to n=3 energy given to electron is E₃−E₁=12.1 eV

Answer

E= EoZ^2{1/n1^2 -1/n2^2}
(Eo=13.6ev
Z=1
n1=1
n2=3
So,E=*13.6×8)/9=12.08~12.1ev

Answer

Given

An electron collides with a hydrogen atom in its ground state and excites it to a state of n = 3.

To find

The amount of energy given to the hydrogen atom in this inelastic collision.

Solution

To calculate the energy given to the hydrogen atom, we need to find the difference in energy between the ground state (n=1) and the excited state (n=3).

The energy levels of a hydrogen atom are given by the formula:

E = -13.6/n^2

where E is the energy of the state and n is the principal quantum number.

Energy of the ground state (n=1)

E1 = -13.6/1^2
= -13.6 eV

Energy of the excited state (n=3)

E3 = -13.6/3^2
= -13.6/9
= -1.511 eV

The energy given to the hydrogen atom in this inelastic collision is the difference in energy between the excited state and the ground state:

ΔE = E3 - E1
= -1.511 eV - (-13.6 eV)
= -1.511 eV + 13.6 eV
= 12.089 eV

Rounding to the nearest tenth, the energy given to the hydrogen atom in this inelastic collision is approximately 12.1 eV.

Therefore, the correct answer is option C) 12.1 eV.

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