Question

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Answer 1

Explanation:

it will help you...........

Answer 2

Given Question:

Calculate the wavelength corresponding to transition from n1 =1 to n2 =4 in Lyman series.

To Find :

Wavelength

Formula used :

$\frac{1}{λ} = R {z}^{2} ( \frac{1}{ {n1}^{2} } - \frac{1}{ {n2}^{2} })$

Here, λ is the wavelength, R is Rydberg's constant and z =1 [H] is Atomic number.

Solution :

The Lyman series is the first line of transition and is hydrogen spectral series.It lies both in emission as well as absorption spectrum.

$\frac{1}{λ} = R {z}^{2} ( \frac{1}{ {n1}^{2} } - \frac{1}{ {n2}^{2} })$

According to the question, we need to find the wavelength so ;

$λ = \frac{1}{R( \frac{1}{ {n1}^{2} } - {\frac{1}{ {n2}^{2} } }) }$

As we know,

$R (constant) = 1.1 \times {10}^{ - 7} {m} ^{-1}$

And,

$\frac{1}{R} = 912 \: Å \:$

Hence,on solving further

$λ = \frac{1}{R( \frac{1}{ {n1}^{2} } - {\frac{1}{ {n2}^{2} } }) } \: \\ \\ λ = \frac{1}{ R \:( \frac{1}{ {1}^{2} } - \frac{1}{ {4}^{2} } )} \\ \\ λ = \frac{1}{R \times \frac{15}{16}} \\ \\ λ = 912 \times \frac{16}{15} Å \:$

$λ = 972.8 \: Å \:$

Therefore,the wavelength of transition in the Lyman series is 972.8 Å.