JEE Main & Advanced Chemistry Structure of Atom / परमाणु संरचना Sample Paper Topic Test - Structure of Atom

Last line of Lyman series for H-atom has wavelength \[{{\lambda }_{1}}\overset{o}{\mathop{A}}\,,\] \[{{2}^{nd}}\] line of Balmer series has wavelength \[{{\lambda }_{2}}\overset{o}{\mathop{A}}\,\] then

A) \[\frac{16}{{{\lambda }_{1}}}=\frac{9}{{{\lambda }_{2}}}\]

B) \[\frac{16}{{{\lambda }_{2}}}=\frac{9}{{{\lambda }_{1}}}\]

C) \[\frac{4}{{{\lambda }_{1}}}=\frac{1}{{{\lambda }_{2}}}\]

D) \[\frac{16}{{{\lambda }_{1}}}=\frac{3}{{{\lambda }_{2}}}\]

Correct Answer: B

Solution :

[b] \[\frac{1}{{{\lambda }_{1}}}=R{{(1)}^{2}}\left( \frac{1}{{{1}^{2}}}-\frac{1}{{{\infty }^{2}}} \right)\]

and \[\frac{1}{{{\lambda }_{2}}}=R{{(1)}^{2}}\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{4}^{2}}} \right)\]

\[\therefore \] \[{{\lambda }_{1}}=\frac{1}{R}\] and \[{{\lambda }_{2}}=\frac{16}{3R}\]

\[\therefore \] \[\frac{16}{{{\lambda }_{2}}}=\frac{3}{{{\lambda }_{1}}}\]