Atoms And Nuclei Question 316
Question: The difference between the longest wavelength line of the Balmer series and shortest wavelength line of the Lyman series for a hydrogenic atom (atomic number Z) equal to $ \Delta \lambda $ . The value of the Rydberg constant for the given atom is :
Options:
A) $ \frac{5}{31}\frac{1}{\Delta \lambda .Z^{2}} $
B) $ \frac{5}{36}\frac{Z^{2}}{\Delta \lambda .} $
C) $ \frac{31}{5}\frac{1}{\Delta \lambda .Z^{2}} $
D) none of these
Show Answer
Answer:
Correct Answer: C
Solution:
- $ \Delta \lambda ={\lambda_{1}}-{\lambda_{2}} $ $ \frac{1}{R[ \frac{1}{{{( 2 )}^{2}}}-\frac{1}{{{( 3 )}^{2}}} ]Z^{2}}=-\frac{1}{R[ 1-0 ]Z^{2}} $ . On solving, $ R=\frac{31}{5}\frac{1}{\Delta \lambda .Z^{2}} $
BETA
