Atoms And Nuclei Question 63
Question: The third line of Balmer series of an ion equivalent to hydrogen atom has wavelength of 108.5 nm. The ground state energy of an electron of this ion will be [RPET 1997]
Options:
A) 3.4 eV
B) 13.6 eV
C) 54.4 eV
D) 122.4 eV
Show Answer
Answer:
Correct Answer: C
Solution:
For third line of Balmer series $ n_{1}=2 $ , $ n_{2}=5 $ \ $ \frac{1}{\lambda }=RZ^{2}[ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} ] $
gives $ Z^{2}=\frac{n_{1}^{2}n_{2}^{2}}{(n_{2}^{2}-n_{1}^{2})\lambda R} $
On putting values Z = 2
From $ E=-\frac{13.6Z^{2}}{n^{2}}=\frac{-13.6{{(2)}^{2}}}{{{(1)}^{2}}}=-54.4,eV $
BETA
