If the wavelength of the n^ th line of Lyman series is equal to the de-Broglie wavelength of electron in initial orbit of a hydrogen like element (Z=11) . Find the value of n .

Step-by-step solution

n^ th line of Lyman series means transition from (n+1) th state to first state. aligned & 1 =R Z^2 [1- 1 (n+1)^2 ] & de-Broglie wavelength, = h m v = h r m v r = (2 )(h r) (n+1) h = 2 r (n+1) & 1 = (n+1) 2 r & aligned or Equating (i) and (ii), we get ( n+1 2 r )=R Z^2 [ n(n+2) (n+1)^2 ] Now, as aligned & r n^2 Z & r= (n+1)^2 11 r₀ aligned r= (n+1)^2 11 r₀ Substituting in equations (iii), we get 11 2 r₀ = R(11)^2(n)(n+2) (n+1) or (n+1)= (1.09 10^7 )(11)(2 ) (0.529 10⁻¹⁰ ) (n^2+2 n ) Solving this equation we get, n=2