Question

calculate the wave number of second line in balmer series of hydrogen spectrum​

Answer 1

Answer:

Given:-

Hydrogen Spectrum = R=1.096 × 10 ` 7  m  −1

To Find:-

Wavenumber and wavelength of the second line in the Balmer series of hydrogen spectrum

Solution:-

For Balmer series n1 = 2 , for third line n2 = 3, for fourth line n2 = 4

Wavenumber vector V of the third line -

V3 - 2 = R [ 1/n1 - 1/n2]

= 1.096 x 10`7 [ 1/2² - 1/3² ]

= 1.096 x 10`7 [ 1/4 - 1/9 ]

= 1.096 x 10`7 x 5/36

= 1.5236 x 106m`-1

= 36/ 1.096 x 10`7 x 5

= 656.3 x 10`-9m

= 656.3nm

Now, wavelength of Ha line

λ = 656.3nm

For Hb line n2 = 4

Thus,

R = 10.97 x 106 ( 1/2² -1/4²)

= 10.97 x 10`6 x 3/16

= 2.0568 x 10`6

λ = 16/10.97 x 10`6 x 3

Answer: Wavelength is 656.3nm and wavenumber is 16/10.97 x 10`6 x 3m

Answer 2

Answer:

The wave number for the second line of H- atom of Balmer series is 20564.43 cm-1 and for limiting line is 27419 cm-1.

Step-by-step explanation:

PlZ mark me as a BRAINLIEST.