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Question: The radius of electron orbit and the speed of electron in the ground state of hydrogen atom is \(5.3...

The radius of electron orbit and the speed of electron in the ground state of hydrogen atom is 5.30×1011m5.30 \times 10^{11}m and 2.2×106ms−12.2 \times 10^{6}ms^{- 1} respectively, then the orbital period of this electron in second excited state will be

A

1.21×10−14 s1.21 \times 10^{- 14}\text{ s}

B

1.21×10−12 s1.21 \times 10^{- 12}\text{ s}

C

1.21×10−10 s1.21 \times 10^{- 10}\text{ s}

D

1.21×10−15 s1.21 \times 10^{- 15}\text{ s}

Answer

1.21×10−15 s1.21 \times 10^{- 15}\text{ s}

Explanation

Solution

Here r1=5.30×10−11mr_{1} = 5.30 \times 10^{- 11}m

v1=2.2×106ms−1v_{1} = 2.2 \times 10^{6}ms^{- 1}

In the second excited state,

rn=n2r1,vn=v1nr_{n} = n^{2}r_{1},v_{n} = \frac{v_{1}}{n}

∴r2=4r1=4×5.30×10−11m=2.12×10−10m\therefore r_{2} = 4r_{1} = 4 \times 5.30 \times 10^{- 11}m = 2.12 \times 10^{- 10}m

And

v2=v12=2.2×1062ms−1=1.1×106ms−1v_{2} = \frac{v_{1}}{2} = \frac{2.2 \times 10^{6}}{2}ms^{- 1} = 1.1 \times 10^{6}ms^{- 1}

Since, orbit period

(T)=2πr2v2(T) = \frac{2\pi r_{2}}{v_{2}}

=2×3.14×2.12×10−101.1×106=13.31×10−161.1= \frac{2 \times 3.14 \times 2.12 \times 10^{- 10}}{1.1 \times 10^{6}} = \frac{13.31 \times 10^{- 16}}{1.1}

=1.21×10−15s.= 1.21 \times 10^{- 15}s.