In hydrogen-like atom z - 11- nth line of Lyman series has wavelength - equal to the de-Broglie-s wavelength of an electron in the level from which it originated- What is the value of n-

The correct option is A 24 1 /λ #160; =R z ^ 2 ( 1 / n _ 1 ^ 2 1 / n _ 2 ^ 2 ) #160; 1 /λ #160; =R (11) ^ 2 ( 1 / 1 1 / n ^ 2 ) #160; ...

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Byju's Answer

Standard XII

Physics

Energy Levels

In hydrogen-l...

Question

In hydrogen-like atom (z = 11), $n^{t h}$ line of Lyman series has wavelength $λ$ equal to the de-Broglie's wavelength of an electron in the level from which it originated. What is the value of n?

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Solution

The correct option is A 24
$\frac{1}{λ} = R z^{2} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$

$\frac{1}{λ} = R {(11)}^{2} (\frac{1}{1} - \frac{1}{n^{2}})$

$λ = \frac{h}{m v}$

$λ = \frac{h r}{m v r} = \frac{2 π h r}{n h} = \frac{2 π r}{n}$

$λ = \frac{2 π r}{n} = \frac{π (0.529 \times 10^{- 10}) n^{2}}{n (11)}$

$\frac{1}{λ} = \frac{11}{π (0.529 \times 10^{- 10}) n} = {{1.1 \times 10}^{7} (11)}^{2} (\frac{1}{1} - \frac{1}{n^{2}})$

$n - \frac{1}{n} = 25$

$n \approx 25$

The transition $25 \to 1$ corresponds to the 24th line of Lyman series and hence n = 24.

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In hydrogen-like atom (z = 11), nth line of Lyman series has wavelength λ equal to the de-Broglie's wavelength of an electron in the level from which it originated. What is the value of n?

The correct option is A 241λ=Rz2(1n21−1n22) 1λ=R(11)2(11−1n2) λ=hmv λ=hrmvr=2πhrnh=2πrn λ=2πrn=π(0.529×10−10)n2n(11) 1λ=11π(0.529×10−10)n=1.1×107(11)2(11−1n2) n−1n=25 n≈25The transition 25→1 corresponds to the 24th line of Lyman series and hence n = 24.

In hydrogen-like atom (z = 11), $n^{t h}$ line of Lyman series has wavelength $λ$ equal to the de-Broglie's wavelength of an electron in the level from which it originated. What is the value of n?