An electron in a hydrogen atom emits wavelengths during the transitions given below. Arrange these wavelengths in ascending order:

(A) from n = 3 to n = 2
(B) from n = 2 to n = 1
(C) from n = ∞ to n = 1
(D) from n = ∞ to n = 3
(E) from n = 4 to n = 1

Choose the correct answer from the options given below:

(C), (E), (B), (A), (D)

The correct answer is Option (4) → (C), (E), (B), (A), (D)

Given:

Electron transitions in hydrogen atom:

(A) n = 3 → n = 2

(B) n = 2 → n = 1

(C) n = ∞ → n = 1

(D) n = ∞ → n = 3

(E) n = 4 → n = 1

Concept:

Energy difference, $E = 13.6\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$ eV

Since $E \propto \frac{1}{\lambda}$, greater energy corresponds to smaller wavelength.

Calculations:

(A) $E_A = 13.6\left(\frac{1}{2^2} - \frac{1}{3^2}\right) = 13.6\left(\frac{5}{36}\right) = 1.89\,\text{eV}$

(B) $E_B = 13.6\left(\frac{1}{1^2} - \frac{1}{2^2}\right) = 13.6\left(\frac{3}{4}\right) = 10.2\,\text{eV}$

(C) $E_C = 13.6\left(\frac{1}{1^2} - \frac{1}{\infty^2}\right) = 13.6\,\text{eV}$

(D) $E_D = 13.6\left(\frac{1}{3^2} - \frac{1}{\infty^2}\right) = 13.6\left(\frac{1}{9}\right) = 1.51\,\text{eV}$

(E) $E_E = 13.6\left(\frac{1}{1^2} - \frac{1}{4^2}\right) = 13.6\left(\frac{15}{16}\right) = 12.75\,\text{eV}$

Energy order (ascending):

(D) < (A) < (B) < (E) < (C)

Hence, wavelength order (ascending):

(C) < (E) < (B) < (A) < (D)

Final Answer: (C), (E), (B), (A), (D)