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  • Need explanation for: - Chemical Bonding and Molecular Structure - JEE Main

The wavelength of the radiation emitted, when in a hydrogen atom electron falls from infinity to stationary state 1, would be \left ( Rydberg\: constant \: = 1.097 \times 10^{7}m^{-1}\right )

  • Option 1)

    91\: nm

  • Option 2)

    192 \: nm

  • Option 3)

    406\: nm

  • Option 4)

    9.1\times 10^{-8}nm

 

Answers (2)

best_answer

As we learnt in 

Line Spectrum of Hydrogen like atoms -

\frac{1}{\lambda }= RZ^{2}\left ( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right )

 

- wherein

Where R is called Rhydberg constant, R = 1.097 X 107 , Z is atomic number

n1= 1,2 ,3….

n2= n1+1, n1+2 ……

 

\frac{1}{\lambda }= R\left ( \frac{1}{{n_{1}}^{2}} -\frac{1}{{n_{2}}^{2}}\right )

\frac{1}{\lambda }= 1.097 \times 10^{7}m^{-1}\left ( \frac{1}{1^{2}}-\frac{1}{\infty ^{2}} \right )

\lambda = 91\times 10^{-9}m= 91nm


Option 1)

91\: nm

Correct option

Option 2)

192 \: nm

Incorrect option

Option 3)

406\: nm

Incorrect option

Option 4)

9.1\times 10^{-8}nm

Incorrect option

Posted by

Aadil

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