De-Broglie wavelength of an electron orbiting | vedclass

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Question

(B) According to Bohr's quantization condition,the circumference of the orbit is an integer multiple of the De-Broglie wavelength: $2 \pi r_n = n \lam

Vedclass · Accepted Answer

$0.67$

Vedclass · Answer

(B) According to Bohr's quantization condition,the circumference of the orbit is an integer multiple of the De-Broglie wavelength: $2 \pi r_n = n \lambda$.The radius of the $n$-th orbit is given by $r_n = a_0 \frac{n^2}{Z}$,where $a_0 = 0.052 \ nm$ and $Z = 1$ for hydrogen.For $n=2$,the radius is $r_2 = 0.052 	imes \frac{2^2}{1} = 0.052 	imes 4 = 0.208 \ nm$.Substituting this into the quantization condition: $2 \pi (0.208) = 2 \lambda$.Solving for $\lambda$: $\lambda = \pi 	imes 0.208 \approx 3.14159 	imes 0.208 \approx 0.653 \ nm$.Rounding to the nearest given option,the value is $0.67 \ nm$.

De-Broglie wavelength of an electron orbiting in the $n=2$ state of a hydrogen atom is close to (Given Bohr radius $= 0.052 \ nm$) (in $nm$)

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