If the de Broglie wavelength of the electron | vedclass

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(D) According to the Bohr model,the circumference of the $n^{th}$ orbit is given by $2 \pi r_n = n \lambda$,where $\lambda$ is the de Broglie waveleng

Vedclass · Accepted Answer

$0.75$

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(D) According to the Bohr model,the circumference of the $n^{th}$ orbit is given by $2 \pi r_n = n \lambda$,where $\lambda$ is the de Broglie wavelength. Substituting the expression for the radius of the $n^{th}$ orbit,$r_n = a_0 \frac{n^2}{Z}$,we get: $2 \pi \left( a_0 \frac{n^2}{Z} \right) = n \lambda$ $\lambda = \frac{2 \pi a_0 n^2}{n Z} = 2 \pi a_0 \frac{n}{Z}$ Given that $\lambda = 1.5 \pi a_0$,we equate the two expressions: $2 \pi a_0 \frac{n}{Z} = 1.5 \pi a_0$ $\frac{n}{Z} = \frac{1.5 \pi a_0}{2 \pi a_0} = \frac{1.5}{2} = 0.75$

If the de Broglie wavelength of the electron in the $n^{th}$ Bohr orbit in a hydrogenic atom is equal to $1.5 \, \pi \, a_0$ ($a_0$ is Bohr radius),then the value of $n/Z$ is:

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