Publish On: 2019-03-21

# Peter

Total Post: 504

## Question: Calculate the longest and shortest wavelengths in the Balmer series of Hydrogen atom. Given Rydberg constant = 1.0987 × 10^{7} m^{-1}

Reply On: 2013-10-10

# lala singh

Total Post: 22

## ANS: Calculate the longest and shortest wavelengths in the Balmer series of Hydrogen atom. Given Rydberg constant = 1.0987 × 10^{7} m^{-1}

The wavelength () of different spectral lines of Balmer series is given by

1/ = R [1/2

Longest wavelength is of Ha line or 1

Therefore 1/ = 1.097 × 10

= 1.097 × 10

= [36/ (5 × 1.097 × 10

For shortest wavelength, n

Therefore, 1/ = 1.097 × 10

= (1.097 × 10

= (4/1.097 × 10

= 4 × 10

= 3646 Å

1/ = R [1/2

^{2}– 1/n^{2}_{2}]Longest wavelength is of Ha line or 1

^{st}line of the series. For which n_{2}= 3Therefore 1/ = 1.097 × 10

^{7}[1/2^{2}– 1/3^{2}]= 1.097 × 10

^{7}× 5/36= [36/ (5 × 1.097 × 10

^{7})] ÅFor shortest wavelength, n

_{2}= ∞Therefore, 1/ = 1.097 × 10

^{7}[1/2^{2}– 1/∞^{2}]= (1.097 × 10

^{7})/ 4= (4/1.097 × 10

^{7}) m= 4 × 10

^{10}Å /1.097 × 10^{7}= 3646 Å

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