Question

The distance between electron and proton in a hydrogen atom is 0.53 A° . Find the speed and frequency of revolution of electron round the proton .
( mass of electron = 9.1 × 10^-31 kg )

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Answer 1

Answer:

centripetal force = electrostatic force of attraction

mv^2/r = kq^2 / r^2

v^2 = mke^2 / r

v = √[9 x 10^-31 x 9 x 10^9 x (1.6 x 10^-19)^2]/0.53 x 10-10

v = (9 x 1.6 x 10^-11 x 10^-20) / 10^-6 √53

v = 9 x 1.6 x 10^-25 / √53

v = 1.97 x 10^-25.

now, for frequency

omega = velocity x radius = 2 πf

so, f = (1.97 x 10^-25 x 0.53 x 10^-10 ) / 2π

= 0.1668 x 10^-35 .

hope it help ☺

Answer 2

Given :

➨ Radius of circular path = 0.53A°

➨ Mass of electron = 9.1×10$^{-31}$kg

To Find :

⟹ Velocity of electron.

⟹ Frequency of revolution.

SoluTion :

➳ First we have to find out principle quantum number of the orbit.

➳ Atomic number of H (Z) = 1

✴ Radius of stationary state of hydrogen-like species :

$\longrightarrow\tt\:r_n=\dfrac{n^2}{z}\times 0.53\:A\degree\\ \\ \longrightarrow\tt\:0.53\:A\degree=\dfrac{n^2}{Z}\times 0.53\:A\degree\\ \\ \longrightarrow\tt\:n^2=\dfrac{0.53}{0.53}=1\\ \\ \longrightarrow\bf\:n=1$

✴ Velocity of electron in nth shell of hydrogen-like species :

$\longrightarrow\tt\:v_n=\dfrac{Z}{n}\times 2.18\times 10^{6}\:ms^{-1}\\ \\ \longrightarrow\tt\:v_{n=1}=\dfrac{1}{1}\times 2.18\times 10^6\\ \\ \longrightarrow\underline{\boxed{\bf{v_{n=1}=2.18\times 10^{6}\:ms^{-1}}}}$

✴ Frequency of revolution :

$\longrightarrow\bf\:Velocity=\dfrac{Distance}{Time\:period}\\ \\ \longrightarrow\sf\:\dfrac{1}{Time\:period}=\dfrac{Velocity}{Distance}\\ \\ \longrightarrow\bf\:Frequency=\dfrac{V_{n=1}}{2\pi\times r_{n=1}}\\ \\ \longrightarrow\tt\:f=\dfrac{2.18\times 10^6}{2\pi\times 0.53\times 10^{-10}}\\ \\ \longrightarrow\tt\:f=\dfrac{2.18}{2\pi\times 0.53}\times 10^{16}\\ \\ \longrightarrow\underline{\boxed{\bf{f=6.54\times 10^{15}\:Hz}}}$