Atomic Structure Question 373
Question: The wavelength of the radiation emitted, when in a hydrogen atom electron falls from infinity to stationary state 1, would be (Rydberg constant $ =,1.097,\times ,10^{7}{m^{-1}} $ ) [AIEEE 2004]
Options:
A) 406 nm
B) 192 nm
C) 91 nm
D) $ 9.1,\times ,{10^{-8}}nm $
Show Answer
Answer:
Correct Answer: C
Solution:
$ \frac{1}{\lambda }=R,[ \frac{1}{n_1^{2}}-\frac{1}{n_2^{2}} ] $
$ \frac{1}{\lambda }=1.097\times 10^{7}{m^{-1}}[ \frac{1}{1^{2}}-\frac{1}{{{\infty }^{2}}} ] $
$ \therefore $ $ \lambda =91\times {10^{-9}}m $ We know $ {10^{-9}}=1,nm $ So $ \lambda =91nm $
BETA
