Question

Electrons in a certain energy level n = $ {{n}_{1}} $ , can emit 3 spectral lines. When they are in another energy level, n - $ {{n}_{2}} $ , they can emit 6 spectral lines. The orbital speed of the electrons in the orbits are in the ratio

• A. 4 : 3
• B. 3 : 4
• C. 2 : 1
• D. 1 : 2

Answer 1

Answer: (A)

Number of emitted spectral lines $ N=\frac{n(n-1)}{2} $ 1st case $ N=3 $ $ \therefore $ $ 3=\frac{{{n}_{1}}({{n}_{1}}-1)}{2} $ $ \Rightarrow $ $ n_{1}^{2}-{{n}_{1}}-6=0 $ $ \Rightarrow $ $ ({{n}_{1}}-3)({{n}_{1}}+2)=0 $ $ {{n}_{1}}=3,{{n}_{1}}=-2 $ Negative value of n^ is not possible $ \therefore $ $ {{n}_{1}}=3 $ 2nd case $ N=6 $ Again, $ 6=\frac{{{n}_{2}}({{n}_{1}}-1)}{2} $ $ \Rightarrow $ $ n_{2}^{2}-{{n}_{2}}-12=0 $ $ \Rightarrow $ $ ({{n}_{2}}-4)({{n}_{2}}+3)=0 $ $ {{n}_{2}}=4,{{n}_{2}}=-3 $ Again, as $ {{n}_{2}} $ is always positive $ \therefore $ $ {{n}_{2}}=4 $ Velocity of electron $ v=\frac{Z{{e}^{2}}}{2{{\varepsilon }_{0}}hn} $ $ \therefore $ $ \frac{{{v}_{1}}}{{{v}_{2}}}=\frac{{{n}_{2}}}{{{n}_{1}}} $ $ \Rightarrow $ $ \frac{{{v}_{1}}}{{{v}_{2}}}=\frac{4}{3} $