The radius of first Bohr's orbit in H-atoms is $r_{1}$ . The corresponding wavelength of an electron in 2nd orbit is:

$6 π r_{1}$

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$4 π r_{1}$

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$2 π r_{1}$

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$3 π r_{1}$

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Solution

The correct option is B
$4 π r_{1}$

From Bohr's postulates, we know

Wavelength = $\frac{c i r c u m f e r e n c e}{N u m b e r o f w a v e s}$

Number of waves in nth orbit = $n$

Therefore,

$λ$ = $\frac{2 π r_{n}}{n}$

& we know

$r_{n}$ = $r_{1} (n^{2})$

$r_{2}$ = $r_{1} (2^{2})$

= $4 r_{1}$

$\Rightarrow$ $λ$ = $\frac{2 π \times 4 r_{1}}{2}$

$\Rightarrow$ $λ$ = $4 π r_{1}$

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Similar questions

The radius of first Bohr's orbit in H-atoms is $r_{1}$ . The corresponding wavelength of an electron in 2nd orbit is:

r_{0}

H -

2^{n d}

r_{1}

r_{0}

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