A particular hydrogen-like ion emits radiatio | vedclass

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(A) The frequency of emitted radiation is given by $
u = R c Z^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} ight)$.For the transition $n=2$ to $n=1$:

Vedclass · Accepted Answer

$32$

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(A) The frequency of emitted radiation is given by $
u = R c Z^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} ight)$.For the transition $n=2$ to $n=1$:$
u_1 = K \left( \frac{1}{1^2} - \frac{1}{2^2} ight) = K \left( 1 - \frac{1}{4} ight) = K \left( \frac{3}{4} ight) = 3 	imes 10^{15} \,Hz$.For the transition $n=3$ to $n=1$:$
u_2 = K \left( \frac{1}{1^2} - \frac{1}{3^2} ight) = K \left( 1 - \frac{1}{9} ight) = K \left( \frac{8}{9} ight)$.Taking the ratio:$\frac{
u_2}{
u_1} = \frac{K(8/9)}{K(3/4)} = \frac{8}{9} 	imes \frac{4}{3} = \frac{32}{27}$.Therefore, $
u_2 = \frac{32}{27} 	imes 3 	imes 10^{15} \,Hz = \frac{32}{9} 	imes 10^{15} \,Hz$.Comparing this with $\frac{x}{9} 	imes 10^{15} \,Hz$, we get $X = 32$.

$A$ particular hydrogen-like ion emits radiation of frequency $3 \times 10^{15} \,Hz$ when it makes a transition from $n=2$ to $n=1$. The frequency of radiation emitted in a transition from $n=3$ to $n=1$ is $\frac{x}{9} \times 10^{15} \,Hz$, where $X = \text{ . . . . . . }$.

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