What is the energy emitted when h atom de excites from x level to x-1 level. Show that for large frequency this equals the classic frequency of revolution

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$\frac{1}{λ} = R \{\frac{1}{(x - 1)^{2}} - \frac{1}{x^{2}}\} \frac{f}{c} = R \{\frac{1}{(x - 1)^{2}} - \frac{1}{x^{2}}\} f = c R \{\frac{1}{(x - 1)^{2}} - \frac{1}{x^{2}}\} f = c R \{\frac{(2 x - 1)}{x^{2} (x - 1)^{2}}\} a n d R = \frac{m e^{4}}{8 \in^{2} \times c h^{3}} f = \frac{c \times m e^{4}}{8 \in^{2} \times c h^{3}} \{\frac{(2 x - 1)}{x^{2} (x - 1)^{2}}\} (f o r x > > 1 (2 x - 1) = 2 x a n d (x - 1) = x) f = \frac{c \times m e^{4}}{8 \in^{2} \times c h^{3}} \{\frac{2 x}{x^{4}}\} = \frac{c \times m e^{4}}{8 \in^{2} \times c h^{3}} \{\frac{2}{x^{3}}\} = \frac{m e^{4}}{4 \in^{2} \times h^{3}} \{\frac{1}{x^{3}}\} . . 1 F r e q u e n c y o f r e v o l u t i o n o f e l e c t r o n f = \frac{V}{2 π r} w h e r e r = \frac{\in x^{2} h^{2}}{π m e^{2}} a n d u \sin g m v r = \frac{x h}{2 π} s o v = \frac{x h}{2 π m r} f_{e l e c t r o n} = \frac{\frac{x h}{2 π m r}}{2 π \times \frac{\in x^{2} h^{2}}{π m e^{2}}} = \frac{m e^{4}}{4 \in^{2} \times h^{3} x^{3}} . . . 2 f r o m e q n 1 a n d 2 b o t h t h e f r e q u e n c y i s e q u a l . R e g a r d s$

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