Question

The radius of the second Bohr orbit for hydrogen atom is: (Planck's Const. $h =6 . 6262 \times 10^{-34}\, Js$; mass of electron $=9 . 1091 \times 10^{-31}\, Kg$ charge of electron $e =1 .60210 \times 10^{-19} C$; permittivity of vacuum $\left.\left(\epsilon_{0}\right)=8 \cdot 854185 \times 10^{-12}\, Kg ^{-1}\, m ^{-3} A ^{2}\right)$

• A. $0.529\, ??
• B. $2.12\, ??
• C. $1.65\, ??
• D. $4.76\, ??

Answer 1

The Correct Option is B

To find the radius of the second Bohr orbit for a hydrogen atom, we use the formula for the radius of the n-th orbit in a hydrogen atom:

r_n = \frac{n^2 \cdot h^2}{4 \pi^2 \cdot m \cdot e^2 \cdot \epsilon_0}

Where:

• n is the principal quantum number (2 for the second orbit).
• h = 6.6262 \times 10^{-34}\, Js is Planck's constant.
• m = 9.1091 \times 10^{-31}\, Kg is the mass of the electron.
• e = 1.60210 \times 10^{-19}\, C is the charge of the electron.
• \epsilon_0 = 8.854185 \times 10^{-12}\, Kg^{-1}\, m^{-3}\, A^2 is the permittivity of vacuum.

Substitute the values into the formula:

r_2 = \frac{(2)^2 \cdot (6.6262 \times 10^{-34})^2}{4 \cdot (\pi)^2 \cdot (9.1091 \times 10^{-31}) \cdot (1.60210 \times 10^{-19})^2 \cdot (8.854185 \times 10^{-12})}

Calculate the terms step-by-step:

• Calculate the numerator: (2)^2 \cdot (6.6262 \times 10^{-34})^2 \approx 1.759 \times 10^{-66}
• Calculate the denominator: 4 \cdot \pi^2 \cdot (9.1091 \times 10^{-31}) \cdot (1.60210 \times 10^{-19})^2 \cdot (8.854185 \times 10^{-12}) \approx 8.209 \times 10^{-67}

Now divide the numerator by the denominator:

r_2 \approx \frac{1.759 \times 10^{-66}}{8.209 \times 10^{-67}} \approx 2.12 \, \text{\AA}

Therefore, the radius of the second Bohr orbit for a hydrogen atom is approximately 2.12 \, \text{\AA}, which matches the correct answer option.