JEE MAIN - Physics (2023 - 25th January Morning Shift - No. 21)
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 $$\frac{20}{x}\lambda_0$$. The value of $$x$$ is _____________.
Respondre
27
Explicació
Second excited state $$ \to $$ first excited state , $n=3$ to $n=2$
$\frac{1}{\lambda_{0}}=R\left(\frac{1}{4}-\frac{1}{9}\right)=\left(\frac{5 R}{36}\right)$
Third excited state $$ \to $$ second orbit , $n=4$ to $n=2$
$\frac{1}{\lambda}=R\left(\frac{1}{4}-\frac{1}{16}\right)=\left(\frac{3}{16} R\right)$
Taking ratio of (1) and (2)
$\frac{\lambda}{\lambda_{0}}=\frac{5}{36} \times \frac{16}{3}=\left(\frac{20}{27}\right)$
$\lambda=\frac{20}{27} \lambda_{0}$
$x=27$
$\frac{1}{\lambda_{0}}=R\left(\frac{1}{4}-\frac{1}{9}\right)=\left(\frac{5 R}{36}\right)$
Third excited state $$ \to $$ second orbit , $n=4$ to $n=2$
$\frac{1}{\lambda}=R\left(\frac{1}{4}-\frac{1}{16}\right)=\left(\frac{3}{16} R\right)$
Taking ratio of (1) and (2)
$\frac{\lambda}{\lambda_{0}}=\frac{5}{36} \times \frac{16}{3}=\left(\frac{20}{27}\right)$
$\lambda=\frac{20}{27} \lambda_{0}$
$x=27$