For the 'H' atom electron in 2nd bohr orbit, the ratio of radius of or

1.	For the 'H' atom electron in 2nd bohr orbit, the ratio of radius of orbit its to de-Broglie wave length is (A)1(B)2.TE(C)27(D)AB
Answer» $\huge \frak { \underline{ \underline{Explanation : -}}}\\$ As we KNOW that, In nth orbit of hydrogen atom, $\bull \: \bf mv_nr_n = \dfrac{nh}{2\pi}\\$ Now, VELOCITY would be, $\bull \: \bf v_n = \dfrac{nh}{2\pi mr_n} \\$ ALSO, The de-Broglie WAVELENGTH is given by, $: \implies \bf \lambda = \dfrac{h}{mv_n} \\ \\ \\$ $: \implies \bf \lambda = \dfrac{h}{m \bigg( \dfrac{nh}{2\pi mr_n } \bigg) } \\ \\ \\$ $: \implies \bf \lambda = \dfrac{2\pi r_n}{n } \\ \\ \\$ $\large: \implies\boxed{ \bf{\dfrac{r_n}{ \lambda} = \dfrac{n}{2\pi}}} \\ \\ \\$ Hence, the correct answer is option 1 (n/2π) !! ________________