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(C) The de-Broglie wavelength $\lambda$ associated with an electron in an orbit of radius $r$ is given by the relation $2\pi r = n\lambda$,where $n$ i

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will increase.

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(C) The de-Broglie wavelength $\lambda$ associated with an electron in an orbit of radius $r$ is given by the relation $2\pi r = n\lambda$,where $n$ is the principal quantum number.For a hydrogen atom,the radius of the $n^{th}$ orbit is $r_n \propto n^2$.Substituting this into the de-Broglie relation: $2\pi (k n^2) = n\lambda$,which simplifies to $\lambda \propto n$.The ground state corresponds to $n = 1$.The third excited state corresponds to $n = 1 + 3 = 4$.Since the principal quantum number $n$ increases from $1$ to $4$,the de-Broglie wavelength $\lambda$ will increase.

When the electron orbiting in a hydrogen atom in its ground state moves to the third excited state,the de-Broglie wavelength associated with it

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