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Bohrs Derivations

Electrons in ...

Question

Electrons in a certain energy level $n = n_{1}$ can emit $6$ spectral lines. When they are in another energy level $n = n_{2}$ , they can emit $10$ spectral lines. The orbital speed of the electrons in the two orbits are in the ratio of

A

5 : 4

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B

4 : 3

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C

3 : 2

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D

6 : 5

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Solution

The correct option is A $5 : 4$
Number of lines in an emission spectrum,
$\frac{n_{1} (n_{1} - 1)}{2} = 6 \Rightarrow n_{1} = 4$
and

$\frac{n_{2} (n_{2} - 1)}{2} = 10 \Rightarrow n_{2} = 5$

The orbital speed of electron, $V_{n} \propto \frac{1}{n}$

$\Rightarrow \frac{V_{1}}{V_{2}} = \frac{n_{2}}{n_{1}} = \frac{5}{4}$

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The only electron in the hydrogen atom resides under ordinary conditions on the first orbit. When energy is supplied, the electron moves to higher energy orbit depending on the amount of energy absorbed. When this electron returns to any of the lower orbits, it emits energy. Lyman series is formed when the electron returns to the lowest orbit while Balmer series is formed when the electron returns to second orbit. Similarly, Paschen, Brackett and Pfund series are formed when electron returns to the third, fourth and fifth orbits from higher energy orbits respectively. The maximum number of lines produced when an electron jumps from nth level to ground level is equal Maximum number of lines produced when an electron jumps from nth level to ground level is equal to.

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