What is the de-Broglie wavelength associated | vedclass

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(B) The de-Broglie wavelength $\lambda$ is given by the relation $\lambda = \frac{h}{mv}$.From Bohr's postulate,$mvr = \frac{nh}{2\pi}$,which implies

Vedclass · Accepted Answer

$9.96 	imes 10^{-8} \, cm$

Vedclass · Answer

(B) The de-Broglie wavelength $\lambda$ is given by the relation $\lambda = \frac{h}{mv}$.From Bohr's postulate,$mvr = \frac{nh}{2\pi}$,which implies $\frac{h}{mv} = \frac{2\pi r}{n}$.Thus,$\lambda = \frac{2\pi r}{n}$.For the $3^{rd}$ orbit of hydrogen,$n = 3$ and the radius $r = n^2 a_0 = 3^2 	imes 0.529 \, \mathring{A} = 9 	imes 0.529 \, \mathring{A}$.Substituting these values: $\lambda = \frac{2\pi 	imes 9 	imes 0.529}{3} = 6\pi 	imes 0.529 \, \mathring{A}$.$\lambda \approx 6 	imes 3.1416 	imes 0.529 \, \mathring{A} \approx 9.97 \, \mathring{A}$.Since $1 \, \mathring{A} = 10^{-8} \, cm$,we have $\lambda = 9.96 	imes 10^{-8} \, cm$.

What is the de-Broglie wavelength associated with the hydrogen electron in its third orbit?

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