Question

Find the number of revolution made by the electron in 3 orbit of hydrogen atom

Answer 1

Answer:

Circumference of the first orbit of hydrogen atom is given by the formula: [a0=4π2me2h2] In a Bohr's model of an atom, when an electron jumps from n = 1 to n = 3

Explanation:

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Answer 2

Answer:

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Answer: 2.2429 × 10¹⁴

$\textsf {Step Explanation:}$

We know that,

According to  bohr's atomic model,

Radius of the nth orbit is given by,

$r_n=\frac{h^2}{4\pi me^2}\times\frac{n^2}{Z}$

Where,

h(Plank's Constant) = 6.62 × 10⁻²⁷ arg-second

m(Mass of electron) = 9.109 × 10⁻²⁸ gram

e(Charge on electron) = 4.808 × 10⁻¹⁰ esu

Putting these values in  above equation, We get a simplified equation,

$r_n=0.529\times10^{-8}\times\frac{n^2}{Z}\;cm\\\;\\r_n=0.529\times10^{-10}\times\frac{n^2}{Z}\;m$

For Velocity of electron in nth orbital,

$V_n=\frac{2\pi e^2}{h}\times\frac{Z}{n}$

On Simplifying the equation

$V_n=2.18\times10^6\times\frac{Z}{n}\;\;m/s$

Now, Revolution per second means Frequency of electron

$f=\frac{V_n}{2\pi r_n}\\\;\\f=\frac{2.18\times10^6\times\frac{Z}{n}}{2\pi\times0.529\times10^{-10}\times\frac{n^2}{Z}}\\\;\\f=\frac{0.6559\times10^{16}}{n^3}$

For Hydrogen atom , Z = 1  and 3rd orbital n = 3

$f=\frac{0.6559\times10^{16}}{3^3}\\\;\\f=0.02429\times10^{16}\\\;\\f=2.2429\times10^{14}\;\text{revolutions per second}$