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  • Stuck here, help me understand: - Atoms And Nuclei - JEE Main-3

The de-Broglie wavelength (\lambda _{B}) associated with the electron orbiting in the second
excited state of hydrogen atom is related to that in the ground state (\lambda _{G}) by

  • Option 1)

    \lambda _{B}= 2\lambda _{G}

  • Option 2)

    \lambda _{B}= 3\lambda _{G}

  • Option 3)

    \lambda _{B}= \lambda _{G}/2

  • Option 4)

    \lambda _{B}= \lambda _{G}/3

 

Answers (2)

best_answer

As we have learned

Bohr quantisation principle -

mvr=\frac{nh}{2\pi } \\2\pi r= n\lambda

- wherein

Angular momentum of an electron in  stationary orbit is quantised.

 

 2\pi r=n\lambda

for 2nd excited state : n=3

r= r_{0}n^{2}=9r_{0}\Rightarrow \lambda _{B}=\frac{2\pi (9r_{0})}{3}

\lambda _{B}= 6\pi r_{0} 

in grounded state ; n=1; r= r_{0} \Rightarrow \lambda _{g}= 2\pi r_{0}

\lambda _{B}=3\lambda g

 

 


Option 1)

\lambda _{B}= 2\lambda _{G}

This is incorrect

Option 2)

\lambda _{B}= 3\lambda _{G}

This is correct

Option 3)

\lambda _{B}= \lambda _{G}/2

This is incorrect

Option 4)

\lambda _{B}= \lambda _{G}/3

This is incorrect

Posted by

Avinash

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