Question

in the spectrum of hydrogen the ratio of the longest wavelength in the lyman series to the longest wavelength in the balmar series​

Answer 1

as Lyman and Balmer series.

For Lyman Series wavelength will be longest when energy the electron has transition from n=2n=2 to n=1n=1 level.

For Balmer  Series wavelength will be longest when  energy the electron has transition from n=3n=3 to n=2n=2 level.

 E_2 -E_1  \propto (- \dfrac{1}{1^2} +  \dfrac{1}{2^2} ) = \dfrac{3}{4}E2−E1 ∝(−121+ 221)=43

 E_3 -E_2  \propto (- \dfrac{1}{3^2} +  \dfrac{1}{2^2} ) = \dfrac{5}{36}E3−E2 ∝(−321+ 221)=365

\dfrac{E_2 -E_1  }{E_3 -E_2 } = \dfrac{\dfrac{3}{4}}{ \dfrac{5}{36}}=\dfrac{27}{5}E3−E2E2−E1 =36543=527

Hence, \dfrac{\lambda _L } {\lambda _B}=\dfrac{5}{27}λBλL=275

Answer 2

4

Explanation:

In spectral series wavelength λ = R (1/n₁² - 1/n₂²)

Here n₁ is the first orbit and n₂ is higher orbit , whereas R is the Rydberg constant .

In case Lyman series , the electron jumps from higher orbit to the First orbit.

In case of maximum wavelength , the higher orbit is infinity and the value of n₁ is 1 .

Thus λ₁ = maximum wavelength of Lyman series = R x 1/1² = R

Similarly in case Balmer series  , the electron jumps fro higher orbit to second orbit .

For maximum wavelength , it jumps from infinity to second orbit .

Thus λ₂ = maximum wavelength for Balmer series = R( 1/2²) = R/4

The ratio = λ₁/λ₂ = 4