Question

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

• A. wave number of the $1^{st}$ line of the lyman series of the $H{e}^{+}$ will be $\frac{108x}{5}$
• B. wave number of the $1^{st}$ line of the lyman series of the $H{e}^{+}$ will be $\frac{36x}{5}$
• C. The wave length of the $2^{nd}$ line of the lyman series of the $H$ atom is will be $\frac{5}{32x}$
• D. The wave length of the $2^{nd}$ line of the lyman series of the $H$ atom is will be $\frac{32}{5x}$

Answer 1

Answer: (A), (C)

The correct options are
A wave number of the $1^{st}$ line of the lyman series of the $H{e}^{+}$ will be $\frac{108x}{5}$
C The wave length of the $2^{nd}$ line of the lyman series of the $H$ atom is will be $\frac{5}{32x}$

We know,

$\overline{\nu }=\frac{1}{\lambda }=R{Z}^2[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$
where $\overline{\nu }$ = wave number of the transition
$n_{1}and{n}_2$ are the orbit number
Z = atomic number
So, for transition from first line of balmer series
$n_1=2and{n}_2=3$
$\overline{\nu }=\frac{1}{\lambda }=R[\frac{1}{2^2}-\frac{1}{3^2}]$ = $x$ = $\frac{5}{36}R$
Wave number of the first line of lyman series for $H{e}^{+}$ transition from $n_2=2$ to $n_1=1$ will be:
$\overline{\nu }=\frac{1}{\lambda }=R\times 2^2[\frac{1}{1^2}-\frac{1}{2^2}]$ = 3R
Or 3R = $\frac{36}{5}\times x\times 3$ = $\frac{108x}{5}$
Wavelength of $2^{nd}$ line of lyman series of H -atom
$\frac{1}{\lambda }=R[\frac{1}{1^2}-\frac{1}{3^2}]$ = $\frac{8R}{9}$
$\frac{1}{\lambda }$ = $\frac{36x}{5}\times$$\frac{8}{9}$
$\lambda$ = $\frac{5}{32x}$