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16/08/2021 11:07 am
Topic starter
When electronic transition occurs 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
(a) \(\frac{12395}{ΔE} \times 10^{-10}\)m
(b) \(\frac{12395}{ΔE} \times 10^{10}\)m
(c) \(\frac{12395}{ΔE} \times 10^{-10}\)cm
(d) \(\frac{12395}{ΔE} \times 10^{10}\)cm
1 Answer
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16/08/2021 11:10 am
Correct answer: (a) \(\frac{12395}{ΔE} \times 10^{-10}\)m
Explanation:
λ = \(\frac{hc}{ΔE}\)
= \(\frac{6.62 \times 10^{-34}Js \times 3 \times 10^8 ms^{-1}}{E \times 1.602 \times 10^{19}J}\)
= \(\frac{12395}{ΔE} \times 10^{-10}\)m