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  • Hydrogen , deuterium <img alt="\left({ }_1 \mathrm{H}^2\right)" src="/latex-

Hydrogen \left({ }_2 \mathrm{H}^1\right), deuterium \left({ }_1 \mathrm{H}^2\right), singly ionized helium \left({ }_2 \mathrm{He}^4\right)^{+}and doubly ionized lithium \left({ }_3 \mathrm{Li}^6\right)^{++} all have one electron around the nucleus. Consider an electron transition from n=2$ to $n=1. If the wave lengths of emitted radiation are \lambda_1, \lambda_2, \lambda_3 and \lambda_4 respectively, then approximately which one of the following is correct?

Option: 1

4 \lambda_1=2 \lambda_2=2 \lambda_3=\lambda_4


Option: 2

\quad \lambda_1=2 \lambda_2=2 \lambda_3=\lambda_4


Option: 3

\lambda_1=\lambda_2=4 \lambda_3=9 \lambda_4


Option: 4

\lambda_1=2 \lambda_2=3 \lambda_3=4 \lambda_4


Answers (1)

best_answer

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

\frac{1}{\lambda_1}=\mathrm{R}(1)^2(3 / 4), \frac{1}{\lambda_2}=\mathrm{R}(1)^2(3 / 4), \frac{1}{\lambda_3}=\mathrm{R}(2)^2(3 / 4), \frac{1}{\lambda_4}=\mathrm{R}(3)^2(3 / 4)

\frac{1}{\lambda_1}=\frac{1}{4 \lambda_3}=\frac{1}{9 \lambda_4}=\frac{1}{\lambda_2}

So, option (3) is correct.

 

 

 

Posted by

rishi.raj

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