Question

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)
A
B​

Answer 1

$\huge \frak { \underline{ \underline{Explanation : -}}}$

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) !!

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