Question

The energy of hydrogen atom in ground state \( (n = 1) \) is \( -13.6 \) eV. To which highest energy-state the hydrogen atom can be made to reach by the photon of energy \( 12.09 \) eV?
OR
What is meant by interference? Write the conditions for constructive and destructive interference.

Hint:

For the atom, add the photon energy to the ground-state energy and match it to \( -13.6/n^2 \). For interference, recall the path-difference conditions \( n\lambda \) and \( (2n-1)\lambda/2 \).

Answer 1

Option 1 (Atoms):
Step 1: The energy of the hydrogen atom in the \( n^{th} \) state is \( E_n = -\dfrac{13.6}{n^2} \) eV. In the ground state \( n = 1 \), so \( E_1 = -13.6 \) eV.
Step 2: When the atom absorbs a photon of energy \( 12.09 \) eV, its energy rises. Final energy \( E_f = E_1 + E_{photon} = -13.6 + 12.09 = -1.51 \) eV.
Step 3: Put \( E_f = -\dfrac{13.6}{n^2} \): \( -\dfrac{13.6}{n^2} = -1.51 \).
Step 4: Solve for \( n^2 \): \( n^2 = \dfrac{13.6}{1.51} = 9.0 \), so \( n = 3 \). Hence the atom can be raised to the third energy state.
\[\boxed{n = 3}\]

Option 2 (Interference):
Step 1: Interference is the redistribution of light energy that occurs when two coherent waves of the same frequency superpose. At some points the resultant intensity is maximum (bright fringe) and at others it is minimum (dark fringe).
Step 2: Condition for constructive interference (bright): path difference \( \Delta = n\lambda \) with \( n = 0,1,2,... \); equivalently phase difference \( \phi = 2n\pi \).
Step 3: Condition for destructive interference (dark): path difference \( \Delta = (2n-1)\dfrac{\lambda}{2} \) with \( n = 1,2,3,... \); equivalently phase difference \( \phi = (2n-1)\pi \).
\[\boxed{\text{Constructive: } \Delta = n\lambda,\quad \text{Destructive: } \Delta = (2n-1)\tfrac{\lambda}{2}}\]