Atoms And Nuclei Question 23
Question: If the wavelength of the first line of the Balmer series of hydrogen is $ 6561\ {AA} $ , the wavelength of the second line of the series should be [CPMT 1984; DPMT 2004]
Options:
A) $ 13122\ {AA} $
B) $ 3280\ {AA} $
C) $ 4860\ {AA} $
D) $ 2187\ {AA} $
Show Answer
Answer:
Correct Answer: C
Solution:
The wavelength of spectral line in Balmer series is given by
$ \frac{1}{\lambda }=R,[ \frac{1}{2^{2}}-\frac{1}{n^{2}} ] $
For first line of Balmer series, n = 3 Þ $ \frac{1}{{\lambda_{1}}}=R,[ \frac{1}{2^{2}}-\frac{1}{3^{2}} ]=\frac{5R}{36} $ ;
For second line n = 4. Þ $ \frac{1}{{\lambda_{2}}}=R,[ \frac{1}{2^{2}}-\frac{1}{4^{2}} ]=\frac{3R}{16} $
$ \frac{{\lambda_{2}}}{{\lambda_{1}}}=\frac{20}{27}\Rightarrow {\lambda_{1}}=\frac{20}{27}\times 6561=4860,{\AA} $
BETA
