Question

$\frac{x}{x+4}$ is the ratio of energies of photons produced due to transition of an electron of hydrogen atom from its (i) third permitted energy level to the second level and (ii) the highest permitted energy level to the second permitted level The value of $x$ will be

Answer 1

Correct Answer: 5

To solve the problem, we need to analyze the energy transitions in a hydrogen atom. The energy levels in a hydrogen atom are given by the formula:

E_n = -\(\frac{13.6}{n^2}\) eV
where \(n\) is the principal quantum number. The energy of a photon emitted during a transition from a higher energy level \(n_1\) to a lower energy level \(n_2\) is:

ΔE = E_{n_2} - E_{n_1} = -\(\frac{13.6}{n_2^2}\) + \(\frac{13.6}{n_1^2}\)
First, let’s find the energy differences for the specified transitions:

• Transition from third (n=3) to second level (n=2):
  ΔE_3→2 = -13.6\(\left(\frac{1}{2^2} - \frac{1}{3^2}\right)\) = 13.6\(\left(\frac{1}{4} - \frac{1}{9}\right)\) = 13.6\(\left(\frac{5}{36}\right)\) = 1.89 eV.
• Transition from the highest level (n=∞) to the second level (n=2):
  ΔE_∞→2 = -13.6\(\left(\frac{1}{2^2} - \frac{1}{∞^2}\right)\) = 13.6\(\left(\frac{1}{4}\right)\) = 3.4 eV.

The ratio of energies given by:

\(\frac{ΔE_{3→2}}{ΔE_{∞→2}} = \frac{1.89}{3.4}\)
Given that this is \(\frac{x}{x+4}\), equating both expressions gives:

\(\frac{x}{x+4} = \frac{1.89}{3.4}\)
Solve for x:

x(3.4) = 1.89(x+4)
3.4x = 1.89x + 7.56
1.51x = 7.56
x ≈ 5