If the radius of first Bohr orbit is 2x, then de Broglie wavelength of electron in 4th orbit is nearly:

Explanation

Bohr's Quantization of Angular Momentum:According to Bohr's model, the angular momentum of an electron in an orbit is quantized:mvr=n2πh Where:m = mass of electronv = velocity of electron r = radius of the orbitn = principal quantum number (orbit number)h = Planck's constant2. de Broglie's Hypothesis:de Broglie proposed that particles have wave-like properties, and the wavelength (λ) of an electron is related to its momentum (mv) by:λ=mvh 3. Combining the Equations:From Bohr's quantization, we can express mv as:mv=2πrnh Substitute this into de Broglie's equation:λ=2πrnh h λ=n2πr 4. Radius of the nth Orbit:The radius of the nth Bohr orbit is proportional to n²:rn =r1 n2where r₁ is the radius of the first Bohr orbit.5. Applying the Given Information:We are given that r₁ = 2x.We want to find the de Broglie wavelength in the 4th orbit (n = 4).The radius of the 4th orbit is:r4 =r1 ∗42=2x∗16=32xNow, substitute n = 4 and r = r₄ into the de Broglie wavelength equation:λ=n2πr4 =42π(32x) λ=16πxTherefore, the de Broglie wavelength of the electron in the 4th orbit is 16πx.