In a hydrogen atom,the wavelength of the ligh | vedclass

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(B) The Rydberg formula for the wavelength of emitted light is given by: $\frac{1}{\lambda} = R\left[ {\frac{1}{n_1^2} - \frac{1}{n_2^2}} \right]$.For

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$\frac{20}{27}{\lambda _0}$

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(B) The Rydberg formula for the wavelength of emitted light is given by: $\frac{1}{\lambda} = R\left[ {\frac{1}{n_1^2} - \frac{1}{n_2^2}} ight]$.For the transition from $n = 3$ to $n = 2$: $\frac{1}{{\lambda _0}} = R\left[ {\frac{1}{2^2} - \frac{1}{3^2}} ight] = R\left[ {\frac{1}{4} - \frac{1}{9}} ight] = \frac{5}{36}R$  ---$(i)$For the transition from $n = 4$ to $n = 2$: $\frac{1}{{\lambda '}} = R\left[ {\frac{1}{2^2} - \frac{1}{4^2}} ight] = R\left[ {\frac{1}{4} - \frac{1}{16}} ight] = \frac{3}{16}R$  ---(ii)Dividing equation $(i)$ by equation (ii):$\frac{\lambda '}{{\lambda _0}} = \frac{5R}{36} 	imes \frac{16}{3R} = \frac{5 	imes 16}{36 	imes 3} = \frac{80}{108} = \frac{20}{27}$.Therefore,$\lambda ' = \frac{20}{27}{\lambda _0}$.

In a hydrogen atom,the wavelength of the light emitted during a transition from the $n = 3$ orbit to the $n = 2$ orbit is ${\lambda _0}$. What will be the wavelength of the light emitted during a transition from the $n = 4$ orbit to the $n = 2$ orbit?

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