Question

In the spectrum of hydrogen atom, the ratio of the longest wavelength in Lyman series to the longest wavelength in the Balmer series is(a) 5/27(b) 1/93(c) 4/9(d) 3/2

Answer 1

(a) 5/27 is the correct option

Wavelengths for lyman series

= 1/R(1-1/4)

=4/3R

And wavelengths for balmer series

=1/R(1/4 -1/9)

=1/R(5/36)

=36/5R

Wavelengths for lyman series/wavelengths for balmer series

=4/3R =5R/36 = 5/27

Therefore, answer is (a)5/27

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Answer 2

Answer:

The ratio of longest wavelength in Lyman series to the longest wavelength in Balmer series is 5/27

Explanation:

For longest wavelength in Lymen series, the difference in two energy levels should be least.

The formula for finding the wavelength is :

$\dfrac{1}{\lambda}=R_H[\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2}]$

$R_H$ is Rydberg constant

For Lyman series, $n_1=1$ and $n_2=2$

$\dfrac{1}{\lambda}=R_H[\dfrac{1}{1}-\dfrac{1}{2^2}]$

$\lambda_1=\dfrac{4}{3R_H}$...........(1)

For Balmer series, $n_1=2$ and $n_2=3$

So, $\dfrac{1}{\lambda}=R_H[\dfrac{1}{2^2}-\dfrac{1}{3^2}]$

$\lambda_2=\dfrac{36}{5R_H}$........(2)

Dividing equation (1) and (2). We get,

$\dfrac{\lambda_1}{\lambda_2}=\dfrac{5}{27}$

Hence, the correct option is (a) 5/27