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(B) According to de Broglie's hypothesis,the wavelength $\lambda$ of an electron in an orbit is given by $\lambda = \frac{h}{mv}$.For an electron in a

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$\lambda _L < \lambda _M < \lambda _N$

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(B) According to de Broglie's hypothesis,the wavelength $\lambda$ of an electron in an orbit is given by $\lambda = \frac{h}{mv}$.For an electron in a hydrogen-like atom,the velocity $v$ is proportional to $\frac{Z}{n}$,where $n$ is the principal quantum number.Thus,$\lambda \propto \frac{1}{v} \propto \frac{n}{Z}$.For a given atom ($Z$ is constant),$\lambda \propto n$.The energy levels $L, M, N$ correspond to principal quantum numbers $n = 2, 3, 4$ respectively.Since $n_L = 2, n_M = 3, n_N = 4$,we have $n_L Therefore,the wavelengths follow the order $\lambda _L However,the question asks for the decreasing order,which is $\lambda _N > \lambda _M > \lambda _L$.

If $\lambda _L, \lambda _M$ and $\lambda _N$ are the wavelengths of an electron in $L, M, N$ energy levels of $H^{-}$ atom respectively,what is their decreasing order?

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