NEET Solved Paper 2016 Question 38
Question: Given the value of Rydberg constant is $ 10^{7}{m^{-1}}, $ the wave number of the last line of the Balmer series in hydrogen spectrum will be :-
Options:
A) $ 0.025\times 10^{4}m^{1} $
B) $ 0.5\times 10^{7}m^{1} $
C) $ 0.25\times 10^{7}m^{1} $
D) $ 2.5\times 10^{7}m^{1} $
Show Answer
Answer:
Correct Answer: C
Solution:
- $ \frac{1}{\lambda }=RZ^{2}( \frac{1}{n_2^{2}}-\frac{1}{n_1^{2}} )=10^{7}\times 1^{2}( \frac{1}{2^{2}}-\frac{1}{{{\infty }^{2}}} ) $
$ \Rightarrow $ wave number $ =\frac{1}{\lambda }=0.25\times 10^{7}{m^{-1}} $
BETA
