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An electron changes its position from orbitn = 2 to the orbit n = 4 of an atom. The wavelength of the emitted radiations is (R = R vdbergs constant)

A)                 \[\mu \]                                              

B)                 \[\frac{1}{2}m{{v}^{2}}=\mu mgs\]

C)                 \[\frac{1}{2}m{{v}^{2}}=\frac{1}{2}\times 10\times {{(10)}^{2}}=500kg.m/s.\]                                    

D)                 \[v=10m/s,\mu =0.5,\,m=10kg,\,g=10m/{{s}^{2}}\]

Correct Answer: B

Solution :

                Wavelength of lines emitted is \[\frac{1}{\lambda }=R\left( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right)\] where R Rydbergs constant. Given,  \[{{n}_{1}}=2,\,{{n}_{2}}=4\] \[\therefore \]  \[\frac{1}{\lambda }=R\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{4}^{2}}} \right)=\frac{3R}{16}\] \[\Rightarrow \]               \[\lambda =\frac{16}{3R}\]