Question

Find the number of revolutions per second made by an electron in 3rd orbit of Hydrogen atom.​

Answer 1

Answer: 2.2429 × 10¹⁴

Step-by-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

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

Answer 2

Answer:

The number of revolutions per second made by an electron in the 3rd orbit of the Hydrogen atom is $2.4\times 10^{14}$.​

Step-by-step explanation:

Now the radius of $3rd$ orbit is

$3^2\times0.529\times10^{-8}\\=4.761\times10^{-8} cm$

We know that

$mvr=\frac{nh}{2\pi }\\v=\frac{nh}{2\pi mr}$

By substituting all the values, we have

$v=7.3\times10^{7} cm/sec$

Now, the number of revolutions per second is

$=\frac{v}{2\pi r} \\=\frac{7.3\times10^{7}}{2\times3.14\times4.761\times10^{-8}}\\ =2.4\times10^{14}$