Question

Calculate the wave number of the spectral line of shortest wavelength appearing in the balmer series of h- spectrum. (r = 1.09 x 107 m 1 )

Answer 1

Answer Is   3649X 10^ -10

Step By step

we know balmer series is ...

then any orbit  electron jump in 2nd orbits

the we call in balmer series

the i have find a wavelength in balmer series for electron is

1/λ = R (1/2² - 1/∞)

  = 1.09 X 10^7 (1/2² - 1/∞)                   ∴  1/∞ = 0

  = 1.09X10^7 ( 1/ 4 -0)

   1/λ = (1.09x10^7)/4

  then we find λ value in this equation

 λ =  4/ 1.09 x10^7

 = 3.64963 10^ -7

so finally  λ = 3649 X 10^ -10

this is a short wavelength in balmer series  thank you