Question #0d44a

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Question #0d44a

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Answer 1 · 2014-11-04T10:24:34

When an electron jumps from a higher energy orbit to lower energy orbit, light is emitted. The enregy of the light emitted is equal to the energy difference between the two orbits.

Here, the energy difference is $3 \cdot 10^{- 19} J$ $3*10^(-19)J$ .

Energy of light is given by $\frac{h \cdot c}{λ}$ $(h*c)/lambda$

So, $3 \cdot 10^{- 19} = \frac{h \cdot c}{λ} \Rightarrow 3 \cdot 10^{- 19} = \frac{6.626 \cdot 10^{- 34} \cdot 3 \cdot 10^{8}}{λ} \Rightarrow λ = \frac{6.626 \cdot 3 \cdot 10^{- 34 + 8 + 19}}{3} \Rightarrow λ = 6.626 \cdot 10^{- 7} m$ $3*10^(-19)=(h*c)/lambda rArr 3*10^(-19)=(6.626*10^(-34)*3*10^8)/lambda rArr lambda=(6.626*3*10^(-34+8+19))/3rArr lambda=6.626*10^(-7)m$

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