Question

The ratio of minimum to maximum wavelengths of radiation that an electron causes in a Bohr’s hydrogen atom is

• A. ½
• B. Zero
• C. 3/4
• D. 27/32

Answer 1

Answer: (C)

3/4

Answer 2

Energy of radiation that corresponds to the energy difference between two energy levels n₁ and n₂ is given as

E = 13.6 $\left( \frac{1}{n_{1}^{2}}\mspace{6mu} - \mspace{6mu}\frac{1}{n_{2}^{2}} \right)$eV

E is minimum when n₁ = 1 & n₂ = 2

⇒ E_min = 13.6 $\left( \frac{1}{1} - \frac{1}{4} \right)\mspace{6mu} eV\mspace{6mu} = \mspace{6mu} 13.6 \times \frac{3}{4}\mspace{6mu} eV$

E is maximum when n₁ = 1 & n₂ = ∞ (the atom is ionized, that is known as ionization energy)

⇒ E_max = 13.6 $\left( 1 - \frac{1}{\infty} \right)$ = 13.6 eV.

∴ $\frac{E_{\min}}{E_{\max}\mspace{6mu} = \mspace{6mu}\frac{3}{4}}$

⇒$\frac{hc/\lambda_{\max}}{hc/\lambda_{\min}\mspace{6mu} = \mspace{6mu}\frac{3}{4}}$

⇒ $\frac{\lambda_{\min}}{\lambda_{\max}\mspace{6mu} = \mspace{6mu}\frac{3}{4}}$