In a hydrogen atom,the de Broglie wavelength | vedclass

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(B) According to Bohr's postulate,the circumference of the orbit is an integral multiple of the de Broglie wavelength: $n \lambda = 2 \pi r_n$. For a

Vedclass · Accepted Answer

$211.6 \pi \; pm$

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(B) According to Bohr's postulate,the circumference of the orbit is an integral multiple of the de Broglie wavelength: $n \lambda = 2 \pi r_n$. For a hydrogen atom,the radius of the $n^{th}$ orbit is given by $r_n = a_0 	imes n^2$,where $a_0 = 52.9 \; pm$ and $Z = 1$. Substituting $n = 2$ for the second orbit: $r_2 = a_0 	imes (2)^2 = 4 a_0$. Now,substituting this into the wavelength formula: $2 \lambda = 2 \pi (4 a_0)$. Therefore,$\lambda = 4 \pi a_0$. Calculating the value: $\lambda = 4 	imes \pi 	imes 52.9 \; pm = 211.6 \pi \; pm$.

In a hydrogen atom,the de Broglie wavelength of an electron in the second Bohr orbit is: [Given that Bohr radius,$a_0 = 52.9 \; pm$]

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