Highly energetic electron are bombarded in a target of an element containing `30` neutrons Tne ratio of nucleus to that of Helium nucleus is `(14)^(1//3)` . Find atomic number of the nucleus`(R = 1.1 xx 10^(7) m^(-1) )`

Answer

(i) `26` (ii) `1.55 xx 10^(18) Hz`

Explanation

(i) from the relation `r prop A^(1//3)`
We have `r_(1)/r_(2)=(A_(1)/A_(2))^(1//3) implies (A_(2)/4)^(1//3)=(14)^(1//3)`
`:. A_(2)=56`
(ii) `Z_(2)-A_(2)-` no. of neutrons `=56-30=26`
`:. F_(ka)=Rc (Z-1)^(2)(1/1^(1)-1/2^(2))=(3Rc)/4 (Z-1)^(2)`
Subtituting the given values of R, c and Z.
We get `f_(ka)=1.55xx10^(18) Hz`