The frequency of the radiation emitted when electron falls from infinity to n=1 state for he+ would be

Asked by arushidabhade | 8th Jul, 2019, 08:11: PM

Expert Answer:

Given:
 
For He2+
 
n1 = 1
 
n2 = ∞
 
We have,
 
1 over straight lambda space equals space RZ squared open parentheses fraction numerator 1 over denominator straight n subscript 1 superscript 2 end fraction minus fraction numerator 1 over denominator straight n subscript 2 superscript 2 end fraction close parentheses

fraction numerator begin display style 1 end style over denominator begin display style straight lambda end style end fraction space equals open parentheses 1.097 cross times 10 to the power of 7 close parentheses cross times open parentheses 2 squared close parentheses open parentheses fraction numerator begin display style 1 end style over denominator begin display style 1 end style end fraction minus fraction numerator begin display style 1 end style over denominator begin display style infinity end style end fraction close parentheses

fraction numerator begin display style 1 end style over denominator begin display style straight lambda end style end fraction space equals space 4.388 space straight s to the power of negative 1 end exponent
 
The frequency of the radiation is 4.388 s-1.

Answered by Varsha | 9th Jul, 2019, 10:47: AM