Question

If the wave number of $1^{st}$line of Balmer series of $H-$atom is $x$, then:

• A. the wavenumber of $1^{st}$ line of Lyman series of the $H{e}^{+}$ ion will be $\frac{108x}{5}$
• B. the wavenumber of $1^{st}$ line of Lyman series of the $H{e}^{+}$ ion will be $\frac{36x}{5}$
• C. the wavelength of $2^{nd}$ line of Lyman series of H-atom is $\frac{5x}{32}$
• D. the wavenumber of $2^{nd}$ line of Lyman series of H-atom is $\frac{32x}{5}$

Answer 1

Answer: (A), (D)

the wavenumber of $1^{st}$ line of Lyman series of the $H{e}^{+}$ ion will be $\frac{108x}{5}$
the wavenumber of $2^{nd}$ line of Lyman series of H-atom is $\frac{32x}{5}$

$V=R{Z}^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

$x=R(\frac{1}{2^2}-\frac{1}{3^2}=)\frac{R5}{36}$

A) $V_1=R\times 2^2(1-\frac{1}{2^2})=3R=\frac{36}{5}\times 3x=\frac{108x}{5}$

C) Wavelength of $2^{nd}$ line of Lyman series of $H-$atom

$\frac{1}{\lambda }=R\times (\frac{1}{1^2}-\frac{1}{3^2})$

$\frac{1}{\lambda }=R\times \frac{8}{9}$

$\frac{1}{\lambda }=\frac{36x}{5}\times \frac{8}{9}$

$\frac{1}{\lambda }=\frac{32}{5}x$