For irreversible expansion of an ideal gas under isothermal condition, the correct option is:

Orbital angular momentum = $\frac{nh}{2\pi }$ $\frac{}{}$

= $\frac{3h}{2\pi }\left(?n=3\right)$

Radius of Bohr’s orbit $R_n=0.529\stackrel{o}{A}\times \frac{n^2}{z}$  

Radius of Bohr’s orbit for hydrogen,   $R_n=0.529\stackrel{o}{A}\times n^2$

For third orbit (R₃) = $0.529\stackrel{o}{A}\times 9$ = r₃ and

Fourth orbit (R₄) = $0.59\stackrel{o}{A}\times 16=r_4$

$\therefore r_4=\frac{16r_3}{9}$  

105.8 × (2/4) = 52.9 pm ⇒ r_Li+ + r_X- = 52.9 × (3²/3) = 158.7 pm

Bohr's theory accounts for the line spectrum of single electron species but Li? has two electrons. Bohr's theory fails to explain splitting of spectral lines in presence of magnetic field i.e. Zeeman effect.

ΔR? = (a? /3) (16 – 9)
ΔR? = (a? /2) (16 – 9)
∴ ΔR? /ΔR? = 2/3