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The wavelength of the radiation emitted is \lambda_0  when an electron jumps from the second excited state to the first excited state of hydrogen atom. If the electron jumps from the third excited state to the second orbit of the hydrogen atom, the wavelength of the radiation emitted will be \frac{20}{x} \lambda_0.The value of x is_______.

Option: 1

27


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

for the hydrogen atoms.

Bohr’s energy is given by

E=-13.6 \times \frac{1}{n^2} \\

 

And \ \ E=\frac{h c}{\lambda}


  For 1^{\text {st }} condition, \frac{h c}{\lambda_0}=13.6\left(\frac{1}{4}-\frac{1}{9}\right)=13.6 \times \frac{5}{36}----(i) \\For 2^{\text {nd }}condition, \frac{h c}{\lambda}=13.6\left(\frac{1}{4}-\frac{1}{16}\right)=13.6 \times \frac{3}{16}----(i i)
Dividing equation (i) by (ii),
\begin{aligned} & \frac{\lambda}{\lambda_0}=\frac{5}{36} \times \frac{16}{3}=\frac{20}{27} \\ & \Rightarrow \lambda=\frac{20}{27} \lambda_0 \\ & \Rightarrow n=27 \end{aligned}

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Riya

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