SATHEE: Atoms And Nuclei Question 319

Atoms And Nuclei Question 319

Question: Taking Rydberg’s constant $R = 1.097 \times 10^{7} m$ , first and second wavelength of Balmer series in hydrogen spectrum is

Options:

A) $2000 \overset{o}{A},, 3000 \overset{o}{A},$

B) $1575 \overset{o}{A},, 2960 \overset{o}{A},$

C) $6529 \overset{o}{A},,, 4280 \overset{o}{A},$

D) $6552 \overset{o}{A},, 4863 \overset{o}{A},$

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Answer:

Correct Answer: D

Solution:

$\frac{1}{λ} = R [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}] .$ For first wavelength, $n_{1}$ $= 2, n_{2} = 3$
$\Rightarrow λ_{1} = 6563 \overset{o}{A}, .$ For second wavelength, n, $= 2, n_{2} = 4$
$\Rightarrow λ_{2} = 4861 \overset{o}{A},$

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Atoms And Nuclei Question 319

Question: Taking Rydberg’s constant R=1.097×107m , first and second wavelength of Balmer series in hydrogen spectrum is

Options:

Answer:

Solution:

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Question: Taking Rydberg’s constant $R = 1.097 \times 10^{7} m$ , first and second wavelength of Balmer series in hydrogen spectrum is