91Ó°ÊÓ

: The formula for the radius of the nth orbit in a hydrogen atom using the Bohr model is given by: r_n = \( n^2 * \frac{4\pi *\varepsilon_0 * h^2}{m * e^2} \) For the first orbit, n = 1. In the problem, we are required to determine the circumference of the first Bohr orbit, which is given by 2πr. Therefore, for the radius of the first Bohr orbit, we have: r_1 = \( \frac{4\pi *\varepsilon_0 * h^2}{m * e^2} \) Where: - ε0 is the permittivity of free space - h is Planck's constant - m is the mass of the electron - e is the elementary charge

: Next, we need to find the circumference, which is given by the formula 2Ï€r. From step 1, for the radius of the first Bohr orbit, r_1: C = 2Ï€r_1

: The de-Broglie wavelength of an electron is given by the formula: λ = \( \frac{h}{mv} \) Since we want to find the de-Broglie wavelength of an electron having the same velocity as the electron in the first Bohr orbit, we can use the relation: v_1 = \( \frac{e^2}{2 * \pi * \varepsilon_0 *h} \) Now, we can substitute this expression for v_1 in the de-Broglie wavelength equation: λ = \( \frac{h}{mv_1} \)

: Now we will find the ratio of the circumference of the first Bohr orbit to the de-Broglie wavelength of electrons having the same velocity as the electron in the first Bohr orbit. Ratio = \( \frac{C}{λ} \) Substitute the values of C and λ from Steps 2 and 3 into the formula and simplify: Ratio = \( \frac{2\pi r_1}{\frac{h}{mv_1}} \) Simplify the ratio expression: Ratio = \( \frac{2\pi r_1 * mv_1}{h} \) Now, we can substitute the expressions for r_1 and v_1 obtained in Steps 1 and 3 and after simplification, we will get: Ratio = \( \frac{2h \pi \left(\frac{4\pi \varepsilon_0 h^2}{me^2}\right)\left(\frac{e^2}{2\pi \varepsilon_0 h}\right)}{mh} \) Upon simplifying the expression, we get: Ratio = 2 Thus, the ratio of the circumference of the first Bohr orbit to the de-Broglie wavelength of electrons having the same velocity is 2:1. Hence, the correct answer is (D) 2:1.