In terms of Rydberg constant `R`, the shortest wavelength in the Balmer series of the hydrogen , atom spestrum will have wavelength

Options

A. 656.3 nm, n= 3
B. 486.1 nm, n=4
C. 410.2 nm, n=5
D. 364.6 nm, n=`oo`

Answer

D

Explanation

The shortest wavelength occurs when an electron makes a transitions from `n=oo to n=2 `state.
` therefore (1)/ ( lamda_(min))= R ((1)/(2^2) -(1)/(oo))=R/4`