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A photon is emitted in transition from \mathrm{n = 4 \: to\: n = 1} level in hydrogen atom. The corresponding wavelength for this transition is (given, \mathrm{h = 4 \times 10^{-15}eVs} ) :

Option: 1

99.3\mathrm{nm}


Option: 2

941\mathrm{nm}


Option: 3

974\mathrm{nm}


Option: 4

94.1\mathrm{nm}


Answers (1)

best_answer

\mathrm{\Delta E=E_{4}-E_{1}}

\mathrm{\frac{hc}{\lambda }=0.85-(-13.6)}

\mathrm{\frac{4\times 10^{-15}\times 3\times 10^{17}}{\lambda_{\left ( nm \right )} }\frac{nm}{s}=12.75}

\mathrm{\lambda =\frac{1200}{12.75}nm}

\mathrm{=94.1nm}

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