Question

using the radii of first three bohr orbit of hydrogen calculate the velocity of an electron in each of these three Orbit

Answer 1

According to bohr in hydrogen atom:

Angular momentum (m × v × r) = nh/2π

⇒ velocity of electron in any orbit is = v = nh/2πmr

here m is the mass of electron = 9.1 × 10⁻³¹ kg

radius of nth orbit = 52.9 pm × n²

h is the planks constant = 6.625 × 10⁻³⁴ J.sec

let us calculate the value of h/2πm which is common to all the velocities so that calculation becomes easier

substituting the values in h/2πm we get

h/2πm = 1.158 × 10⁻⁴

now velocity of the electron in 1st orbit = h/2πmr₁ ( n = 1 for 1st orbit)

where r₁ is the radius of 1st orbit = 52.9 × 10⁻¹² m

we got the h/2πm value and r₁ substituting the values in v₁ expression we get

v₁ = 2.1 × 10⁶m/sec

now velocity of the electron in second orbit = v₁/2 = 1.05 × 10⁶ m/sec

now velocity of the electron in third orbit = v₁/3 = 7 × 10⁵ m/sec

• how v₂ = v₁/2 i would like to explain:
• we know that v₂ = 2h/2πmr₂
• but r₂ = 52.9 pm × 2²
• subsituting r₂ in v₂ 2 in the numerator and 2 from 2² of denominator gets cancelled and the remaining term is
• v₂ = h/(2πm × 52.9 pm × 2)
• but we know that h/(2πm × 52.9) = v₁
• so v₂ = v₁ × 1/2 = v₁/2
• in the similar way v₃ = v₁/3 can be derived.