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Question: If speed of electron in ground state energy level\(2.2 \times 10^{6}\text{ m }\text{s}^{- 1},\) is t...

If speed of electron in ground state energy level2.2×106 m s−1,2.2 \times 10^{6}\text{ m }\text{s}^{- 1}, is then its speed in fourth excited state will be

A

6.8×106 m s−16.8 \times 10^{6}\text{ m }\text{s}^{- 1}

B

8.8×105 m s−18.8 \times 10^{5}\text{ m }\text{s}^{- 1}

C

5.5×105 m s−15.5 \times 10^{5}\text{ m }\text{s}^{- 1}

D

5.5×106 m s−15.5 \times 10^{6}\text{ m }\text{s}^{- 1}

Answer

5.5×105 m s−15.5 \times 10^{5}\text{ m }\text{s}^{- 1}

Explanation

Solution

According to Bohr’s model

v=2Ke2Znhv = \frac{2Ke^{2}Z}{nh}

v∝1nv \propto \frac{1}{n} ∴VAVB=nBnA\therefore\frac{V_{A}}{V_{B}} = \frac{n_{B}}{n_{A}}

Here, VA=2.2×106ms−1V_{A} = 2.2 \times 10^{6}ms^{- 1}

nA=1,nB=4n_{A} = 1,n_{B} = 4n

∴VB=vA×nAnB\therefore V_{B} = v_{A} \times \frac{n_{A}}{n_{B}}

=2.2×106×14=0.55×106=5.5×105ms−1= 2.2 \times 10^{6} \times \frac{1}{4} = 0.55 \times 10^{6} = 5.5 \times 10^{5}ms^{- 1}