De Broglie Wavelength In Bohr Orbits

Question

According to de Broglie's interpretation of stationary states, a standing electron wave exactly fits around each Bohr orbit. For the second orbit of hydrogen, where the radius is 2.12 Å, what is the de Broglie wavelength of the orbiting electron?

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De Broglie explained Bohr's quantisation by requiring that an integral number of electron wavelengths fit around each orbit, expressed as $2\pi r_n = n\lambda$. Rearranging gives the wavelength $\lambda = \dfrac{2\pi r_n}{n}$, which links the wave picture to the orbit geometry. For the second orbit $n = 2$ and $r_2 = 2.12$ Å, so $\lambda = \dfrac{2\pi 	imes 2.12}{2} = \pi 	imes 2.12 \approx 6.66$ Å. The value 3.33 Å is wrong because it forgets the factor $2\pi$ and merely halves the radius. The value 13.3 Å equals the full circumference $2\pi r_2$ and mistakenly omits dividing by $n = 2$. The value 2.12 Å is simply the orbit radius and confuses radius with wavelength. This standing-wave requirement directly reproduces Bohr's angular-momentum rule $m v r = n\hbar$, showing that quantisation is a natural consequence of the electron's wave nature rather than an arbitrary postulate. This directly reflects the NCERT account of de Broglie's standing-wave condition. A plausibility check confirms two complete wavelengths ($2 	imes 6.66 = 13.3$ Å) reproduce the full orbital circumference, exactly as the $n = 2$ standing-wave condition demands.

De Broglie Wavelength In Bohr Orbits

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