The ionization potential of hydrogen atom is 13.6 eV, and λ_H and λ_D denote longest wavelengths in Balmer spectrum of hydrogen and deuterium atoms. respectively. Ignoring the fine and hyperfine structures, the percentage difference $$y = \frac{\lambda_H - \lambda_D}{\lambda_H} \times 100$$​, is closest to

Correct option is C

Solution:

​For the Balmer series, the formula is:

​$$\frac{1}{\lambda} = R^M \left[ \frac{1}{2^2} - \frac{1}{n^2} \right]; \quad R^M = \frac{\mu}{m_e} R^\infty$$ 

For the longest wavelength:

$$\frac{1}{\lambda} = R^M \left[ \frac{1}{2^2} - \frac{1}{3^2} \right]$$ 

For hydrogen atoms, the longest Balmer wavelength can be written as follows:

$$\frac{1}{\lambda_H} = \frac{\mu_H}{m_e} R^\infty \frac{5}{36} = \frac{m_p}{m_p + m_e} R^\infty \frac{5}{36} \Rightarrow \lambda_H = \frac{m_p + m_e}{m_p} \frac{36}{5 R^\infty}$$

Similarly, for deuterium, the longest Balmer wavelength can be written as follows:

$$\frac{1}{\lambda_D} = \frac{\mu_D}{m_e} R^\infty \frac{5}{36} = \frac{2 m_p}{2 m_p + m_e} R^\infty \frac{5}{36} \Rightarrow \lambda_D = \frac{2 m_p + m_e}{2 m_p} \frac{36}{5 R^\infty}$$

$$\frac{\lambda_H - \lambda_D}{\lambda_H} \times 100 = \frac{\frac{m_p + m_e}{m_p} - \frac{2m_p + m_e}{2m_p}}{\frac{m_p + m_e}{m_p}} \times 100 = \left( 1 - \frac{1}{2} \frac{2 \times 1836 m_e + m_e}{1836 m_e + m_e} \right) \times 100 = 0.03\%$$​​​​