The maximum wavelength that would excite an electron from energy level n=1 to n=3 in atomic hydrogen is:

1 n m

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102 n m

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487 n m

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10^{- 7} n m

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Solution

The correct option is B $102 n m$
Energy of $1^{s t}$ orbit $= - 13.6 e v$
$E n e r g y \propto \frac{1}{n^{2}}$
Energy of $3^{r d}$ orbit $= \frac{- 13.6}{9} = - 1.51 e v$
$Δ E = 12.09 e v$
$\frac{h c}{λ} = 12.09 e v$
$λ = 102 n m$

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Similar questions

E = - 2.178 \times 10^{- 18} (\frac{Z^{2}}{n^{2}}) J

n = 1

n = 2

(h = 6.62 \times 10^{- 34} J s a n d c = 3 \times 10^{8} m s^{- 1})

E = - 2.178 \times 10^{- 18} J (\frac{Z^{2}}{n^{2}})

(h = 6.62 \times 10^{- 34} J s a n d c = 3.0 \times 10^{8} {m s}^{- 1})

n = 1

n = 2

(h) = 6.62 \times 10^{- 34} Js

(c) = 3 \times 10^{8} {ms}^{- 1}

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