Question

Consider the Bohr model of hydrogen atom. The speed of an electron in the second orbit (n = 2) is ________ $x 10^6 m/s$ (Round off to 2 decimal places). [Use $h = 6.63 x 10^{-34} Js,\,\, e = 1.6 x 10^{-19} C, \,\,\epsilon _0 = 8.85 x 10^{-12} C^2m^2/N$]

Answer 1

Correct Answer: 1.08

The Bohr model of the hydrogen atom gives us a way to determine the speed of an electron in a given orbit. For an electron in the nth orbit, the speed \( v_n \) is calculated using the formula: \( v_n = \frac{e^2}{2\epsilon_0 h} \times \frac{1}{n} \).
The formula can be rearranged (as derived from resolving the force on the electron in orbit) specifically for hydrogen’s nth orbit, which simplifies as:
\( v_n = \frac{e^2}{2\hspace{0.1em}\epsilon_0\hspace{0.1em}h} \) for \( n = 1\).
For \( n = 2 \), the speed \( v_2 \) can be calculated by dividing the speed in the first orbit by 2: 

\( v_2 = \frac{v_1}{2} = \frac{e^2}{2\hspace{0.1em}\epsilon_0\hspace{0.1em}h \times n} \) 

Substituting \( e = 1.6 \times 10^{-19} \, C \), \( \epsilon_0 = 8.85 \times 10^{-12} \, C^2m^2/N \), \( h = 6.63 \times 10^{-34} \, Js \), and \( n = 2 \):

\( v_2 = \frac{(1.6 \times 10^{-19})^2}{2 \times 8.85 \times 10^{-12} \times 6.63 \times 10^{-34} \times 2} \)

Calculate this:
Step 1: Calculate the numerator: \( (1.6 \times 10^{-19})^2 = 2.56 \times 10^{-38} \) 
Step 2: Calculate the denominator: \( 2 \times 8.85 \times 10^{-12} \times 6.63 \times 10^{-34} \times 2 = 2.34521 \times 10^{-33} \) 
Step 3: Divide the numerator by the denominator: 

\( v_2 = \frac{2.56 \times 10^{-38}}{2.34521 \times 10^{-33}} = 1.091850679 \times 10^{-5} \times 10^6 \, m/s \approx 1.09 \times 10^6 \, m/s \) 

Rounding off to 2 decimal places, the speed of the electron in the second orbit is \( 1.09 \times 10^6 \, m/s \)