If the energy difference between the ground state and excited state of an atom is $4.4 x 10^{- 19} J$ . The wavelength of photon required to produce this transition is
$3.5 x 10^{- 8} m$
$3.5 x 10^{- 7} m$
$4.5 x 10^{- 7} n m$
$4.5 x 10^{- 7} m$

4.5 x 10^{- 7} n m

3.5 x 10^{- 8} m

3.5 x 10^{- 7} m

4.5 x 10^{- 7} m

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Solution

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$Δ E = 4.4 \times 10^{- 19} J$
$Δ E = \frac{h c}{λ}$
$λ = \frac{6.626 \times 10^{- 34} J \times 3 \times 10^{8} m s^{- 1}}{4.4 \times 10^{- 19} J}$
$λ = 4.5 \times 10^{- 7} m$

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If the energy difference between the ground state and excited state of an atom is 4.4x10−19J. The wavelength of photon required to produce this transition is 3.5x10−8m3.5x10−7m4.5x10−7nm4.5x10−7m

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If the energy difference between the ground state and excited state of an atom is $4.4 x 10^{- 19} J$ . The wavelength of photon required to produce this transition is
$3.5 x 10^{- 8} m$
$3.5 x 10^{- 7} m$
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$4.5 x 10^{- 7} m$

$Δ E = 4.4 \times 10^{- 19} J$
$Δ E = \frac{h c}{λ}$
$λ = \frac{6.626 \times 10^{- 34} J \times 3 \times 10^{8} m s^{- 1}}{4.4 \times 10^{- 19} J}$
$λ = 4.5 \times 10^{- 7} m$