Based on the equation ΔE=−2.0×10−18J1n22−1n12, the wavelength of light that must be absorbed to excite hydrogen electron from level n=1 to level n=2 will be (Given: h=6.625×10−34Js,c=3×108ms−1 )

Based on the equation $ΔE = - (2 .0 \times 10^{- 18} J) (\frac{1}{n_{2}^{2}} - \frac{1}{n_{1}^{2}})$ , the wavelength of light that must be absorbed to excite hydrogen electron from level $n = 1$ to level $n = 2$ will be (Given: $h = 6 .625 \times 10^{- 34} Js, c = 3 \times 10^{8} {ms}^{- 1})$

A
$1.325 \times 10^{- 7} m$
B
$1.325 \times 10^{- 10} m$
C
$2.650 \times 10^{- 7} m$
D
$5.300 \times 10^{- 10} m$

Solution:

$ΔE = - (2 .0 \times 10^{- 18} J) (\frac{1}{n_{2}^{2}} - \frac{1}{n_{1}^{2}}) = - (2 .0 \times 10^{- 18} J) (\frac{1}{4} - \frac{1}{1}) = (2 .0 \times 10^{- 18} J) (3 / 4) = 1 .5 \times 10^{- 18} J$

$λ = \frac{hc}{ΔE} = \frac{(6 .625 \times 10^{- 34} Js) (3 \times 10^{8} {ms}^{- 1})}{(1 .5 \times 10^{- 18} J)} = 1 .325 \times 10^{- 7} m$

Solution:

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