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If the radius of first Bohr orbit of hydrogen atom is 'x' then de Broglie wavelength of electron in $3^{rd}$ Orbit is nearly

$(a)\;2\pi x\qquad(b)\;6\pi x\qquad(c)\;9x\qquad(d)\;\large\frac{x}{3}$

1 Answer

$r_n = n^2r_1$
$r_3 = 9r_1 = 9x$
$mvr = \large\frac{h}{2\pi}$
$mv9x = 3\large\frac{h}{2\pi}$
$\large\frac{h}{mv} = 6\pi x$
$\lambda = 6\pi x$
Hence answer is (b)
answered Feb 7, 2014 by sharmaaparna1

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