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Calculate the velocity of electron in Bo...

Calculate the velocity of electron in Bohr's first orbit of hydrogen atom. How many times does the electron go in Bohr's first orbit in one second ?

A

`6.57 xx 10^(5)`

B

`6.57 xx 10^(10)`

C

`6.57 xx 10^(13)`

D

`6.57 xx 10^(15)`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem of calculating the velocity of an electron in Bohr's first orbit of a hydrogen atom and determining how many times the electron goes around this orbit in one second, we can follow these steps: ### Step 1: Calculate the velocity of the electron in the first orbit The formula for the velocity of an electron in the nth orbit of a hydrogen atom is given by: \[ v_n = \frac{Z \cdot e^2}{2 \cdot \epsilon_0 \cdot h} \cdot \frac{1}{n} \] For hydrogen, the atomic number \(Z = 1\) and for the first orbit, \(n = 1\). A more simplified form of this equation is: \[ v_n = \frac{c}{137} \cdot \frac{Z}{n} \] Substituting \(Z = 1\) and \(n = 1\): \[ v_1 = \frac{c}{137} \] Where \(c\) (the speed of light) is approximately \(3 \times 10^8 \, \text{m/s}\). Calculating \(v_1\): \[ v_1 = \frac{3 \times 10^8}{137} \approx 2.18 \times 10^6 \, \text{m/s} \] ### Step 2: Calculate the frequency of the electron in the first orbit The frequency \(f\) of the electron in the nth orbit is given by: \[ f_n = \frac{Z^2 \cdot e^4 \cdot m}{(4 \pi \epsilon_0)^2 \cdot h^3} \cdot \frac{1}{n^3} \] For hydrogen, this can be simplified to: \[ f_n = 6.58 \times 10^{15} \cdot \frac{Z^2}{n^3} \] Substituting \(Z = 1\) and \(n = 1\): \[ f_1 = 6.58 \times 10^{15} \, \text{Hz} \] ### Step 3: Determine how many times the electron goes around the orbit in one second The number of times the electron goes around the orbit in one second is equal to the frequency \(f_1\): \[ \text{Number of revolutions per second} = f_1 = 6.58 \times 10^{15} \, \text{Hz} \] ### Final Answers 1. The velocity of the electron in Bohr's first orbit of hydrogen atom is approximately \(2.18 \times 10^6 \, \text{m/s}\). 2. The electron goes around the Bohr's first orbit approximately \(6.58 \times 10^{15}\) times in one second. ---
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