Solveeit Logo
Login

Question

Question: If an electron in hydrogen atom is revolving in a circular track of radius \(5.3 \times 10^{- 11}m\)...

If an electron in hydrogen atom is revolving in a circular track of radius 5.3×10−11m5.3 \times 10^{- 11}m with a velocity of

2.2×106ms−12.2 \times 10^{6}ms^{- 1}around the proton then the frequency of electron moving around the proton is

A

6.6×1012 Hz6.6 \times 10^{12}\text{ Hz}

B

3.3×1015 Hz3.3 \times 10^{15}\text{ Hz}

C

3.3×1012 Hz3.3 \times 10^{12}\text{ Hz}

D

6.6×1015 Hz6.6 \times 10^{15}\text{ Hz}

Answer

6.6×1015 Hz6.6 \times 10^{15}\text{ Hz}

Explanation

Solution

Frequency of the electron moving around the proton is

υ=velocityofelectron(v)circumference(2πr)\upsilon = \frac{velocityofelectron(v)}{circumference(2\pi r)}

Here, v=2.2×106ms−1v = 2.2 \times 10^{6}ms^{- 1} and r=5.30×10−11mr = 5.30 \times 10^{- 11}m

∴υ=2.2×1062×3.14×5.3×10−11\therefore\upsilon = \frac{2.2 \times 10^{6}}{2 \times 3.14 \times 5.3 \times 10^{- 11}}

υ=6.6×1015Hz\upsilon = 6.6 \times 10^{15}Hz