Question

The value of $n$ for which de Broglie wavelength of $H$ like atom corresponds to $n$th orbit is equal to the wavelength of $n$th line of Lyman series?(Given $Z=11$)

• A. $20$
• B. $25$
• C. $5$
• D. $30$

Answer 1

The correct option is B $25$

As we know,
$2\pi r_n=n{\lambda }_1$
$\therefore {\lambda }_1=\frac{2\pi r_n}{n}=\frac{2\pi \times 0.53\times n^2}{Z\times n}\times {10}^{-10}m$
Also, $\frac{1}{{\lambda }_2}=R_H\cdot Z^2[1-\frac{1}{(n+1{)}^2}]$
$=1.097\times {10}^7\times (11{)}^2\times \left[\frac{n^2+2n}{(n+1{)}^2}\right]$
as ${\lambda }_1={\lambda }_2$
by solving $n=25$