In a hydrogen atom the transition takes place from n=3 and n=2.If the Rydberg constant is 1.09*10^7 metre^-1The wavelength of the emitted radiation is?

According to Rydberg formula,
$\stackrel{-}{\nu }$ (m^-) = 1.09 $\times$10⁷ $\left(\frac{1}{{{\text{n}}_1}^2}-\frac{1}{{{\text{n}}_2}^2}\right)$
$\stackrel{-}{\nu }$= wave number
n₁ = 3
n₂ = 2
R_H (Rydberg constant ) = 1.09$\times$10⁷ m^-1
$\stackrel{-}{\nu }$ = $1.09\times {10}^7 \left(\frac{1}{3^2}-\frac{1}{2^2}\right)$
$\stackrel{-}{\nu }$  = $1.09\times {10}^7\left(\frac{1}{9}-\frac{1}{4}\right)$
$\stackrel{-}{\nu }$ = - 0.151$\times$10⁷ m^-1
It is an emission spectra , hence $\stackrel{-}{\nu }$  is negative.
$\stackrel{-}{\nu }$ = $\frac{1}{\lambda }$
$\lambda$ = 6.622$\times$10^-7 m
Hence, wavelength of emitted radiation is 6.622$\times$10^-7 m.