When electronic transition occur from higher energy state to lower energy state with energy difference equal to -E electron volts- the wavelength of the line emitted is approximately equal to -

The correct option is A 12395/Δ E × 10 ^ 10 mAs we know, λ = #160; hc/Δ E = #160; 6.62 × 10^ 34 Js × 3 × 10^8 ms^ 1/Δ E × 1. ...

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Byju's Answer

Standard X

Chemistry

Bohr's Model of Atom

When electron...

Question

When electronic transition occur from higher energy state to lower energy state with energy difference equal to $Δ$ E electron volts, the wavelength of the line emitted is approximately equal to :

\frac{12395}{Δ E}

\times

^{- 10}

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\frac{12395}{Δ E}

\times

^{10}

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\frac{12395}{Δ E}

\times

^{- 10}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{12395}{Δ E}

\times

^{10}

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Solution

The correct option is A $\frac{12395}{Δ E}$ $\times$ 10 $^{- 10}$ m
As we know,
$λ$ = $\frac{h c}{Δ E}$
= $\frac{6.62 \times 10^{- 34} J s \times 3 \times 10^{8} m s^{- 1}}{Δ E \times 1.602 \times 10^{- 19} J}$
= $\frac{12395}{Δ E}$ $\times$ 10 $^{- 10}$ m

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When electronic transition occur from higher energy state to lower energy state with energy difference equal to ΔE electron volts, the wavelength of the line emitted is approximately equal to :

The correct option is A 12395ΔE × 10−10 mAs we know,λ = hcΔE = 6.62×10−34Js×3×108ms−1ΔE×1.602×10−19J=12395ΔE × 10−10 m

When electronic transition occur from higher energy state to lower energy state with energy difference equal to $Δ$ E electron volts, the wavelength of the line emitted is approximately equal to :

The correct option is A $\frac{12395}{Δ E}$ $\times$ 10 $^{- 10}$ m
As we know,
$λ$ = $\frac{h c}{Δ E}$
= $\frac{6.62 \times 10^{- 34} J s \times 3 \times 10^{8} m s^{- 1}}{Δ E \times 1.602 \times 10^{- 19} J}$
= $\frac{12395}{Δ E}$ $\times$ 10 $^{- 10}$ m