The De Broglie wavelength lambda of an electr | vedclass

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(A) According to Bohr's postulate,the angular momentum of an electron in an orbit is given by $mvr = \frac{nh}{2\pi}$.Rearranging this,we get $2\pi r

Vedclass · Accepted Answer

$4\pi r_1$

Vedclass · Answer

(A) According to Bohr's postulate,the angular momentum of an electron in an orbit is given by $mvr = \frac{nh}{2\pi}$.Rearranging this,we get $2\pi r = \frac{nh}{mv}$.Since the De Broglie wavelength is $\lambda = \frac{h}{mv}$,we can substitute this into the equation to get $2\pi r = n\lambda$,or $\lambda = \frac{2\pi r}{n}$.For the $2^{nd}$ Bohr orbit,$n = 2$.The radius of the $n^{th}$ orbit is given by $r_n = n^2 r_1$,where $r_1$ is the radius of the first Bohr orbit.For $n = 2$,$r_2 = 2^2 r_1 = 4r_1$.Substituting these values into the wavelength formula: $\lambda = \frac{2\pi (4r_1)}{2} = 4\pi r_1$.

The De Broglie wavelength $\lambda$ of an electron in the $2^{nd}$ Bohr orbit is:

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