Question

the first line in balmer series corresponds to n1=2 n2=3 and the limiting lines corresponds to n1=2 n2=∞ calculate the wavelength of first and limiting lines in balmer series
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Answer 1

Given:

The first line in balmer series corresponds to n1=2 n2=3 and the limiting lines corresponds to n1=2 n2=∞  

To find:

Calculate the wavelength of first and limiting lines in balmer series

Solution:

We use the formula,

1/λ = R (1/n₁² - 1/n₂²)

where, R = 1.097  × 10^7 m^{-1}

From given we have,

a) n1 = 2 n2 = 3

 1/λ = R (1/2² - 1/3²)

λL = 36/5R

substituting the value of R, we get,

λL = 656.3 nm

Longest wavelength is emitted in Balmer series if the transition of electron takes place from n2  = 3 to n1 = 2

b) n1 = 2 n2 = ∞

 1/λ = R (1/2² - 1/∞²)

λs = 4/R

substituting the value of R, we get,

λs = 364.6 nm

Smallest wavelength is emitted in Balmer series if the transition of electron takes place from n2  = ∞  to n1 = 2