Given- R-1-097 - 107 m-1- the wavelength of K- line of silver z - 47 being take b-1 for K series

The correct option is C 0.574 A^0 1λ=R(z 6)^2(1n_1^2 1n_2^2) 1λ #160;=1.097× 10^7(47 1 )^2 ( 11 12^2 ) λ =0.574 A^0 ...

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Byju's Answer

Standard XII

Physics

Hydrogen Spectra

Given: R=1....

Question

Given: $R = 1.097 \times 10^{7} m^{- 1}$ , the wavelength of $K_{α}$ line of silver $(z = 47)$ being (take $b = 1$ for K series)

0.674

A^{0}

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0.574

A^{0}

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0.774

A^{0}

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0.287

A^{0}

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Solution

The correct option is C $0.574$ $A^{0}$
$\frac{1}{λ} = R (z - 6)^{2} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$
$\frac{1}{λ} = 1.097 \times 10^{7} (47 - 1)^{2} (\frac{1}{1} - \frac{1}{2^{2}})$
$λ = 0.574 A^{0}$

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Given: R=1.097×107m−1, the wavelength of Kα line of silver (z=47) being (take b=1 for K series)

The correct option is C 0.574A01λ=R(z−6)2(1n21−1n22)1λ=1.097×107(47−1)2(11−122)λ=0.574 A0

Given: $R = 1.097 \times 10^{7} m^{- 1}$ , the wavelength of $K_{α}$ line of silver $(z = 47)$ being (take $b = 1$ for K series)

The correct option is C $0.574$ $A^{0}$
$\frac{1}{λ} = R (z - 6)^{2} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$
$\frac{1}{λ} = 1.097 \times 10^{7} (47 - 1)^{2} (\frac{1}{1} - \frac{1}{2^{2}})$
$λ = 0.574 A^{0}$