MCQ
When electronic transition occurs from higher energy state to lower energy state with energy difference equal to $\Delta E$ electron volts, the wavelength of the line emitted is approximately equal to
  • $\frac{{12395}}{{\Delta E}}\, \times \,{10^{ - 10}}\,m$
  • B
    $\frac{{12395}}{{\Delta E}}\, \times \,{10^{ 10}}\,m$
  • C
    $\frac{{12395}}{{\Delta E}}\, \times \,{10^{ - 10}}\,cm$
  • D
    $\frac{{12395}}{{\Delta E}}\, \times \,{10^{ 10}}\,cm$

Answer

Correct option: A.
$\frac{{12395}}{{\Delta E}}\, \times \,{10^{ - 10}}\,m$
a
$\lambda=\frac{\mathrm{hc}}{\Delta \mathrm{E}}=\frac{6.62 \times 10^{-34} \,\mathrm{Js} \times 3 \times 10^{8} \,\mathrm{ms}^{-1}}{\mathrm{E} \times 1.602 \times 10^{-19} \,\mathrm{J}}$

$=12395 \times 10^{-10} \,\mathrm{m}$

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