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The radius of the second orbit of an electron in hydrogen atom is 2.116 $A^{0}$ . The de-Broglie wavelength associated with this electron in this orbit would be

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$6.64 A^{0}$

$1.058 A^{0}$

$20116 A^{0}$

$13.28 A^{0}$

answer is A.

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Detailed Solution

$mvr = \frac{2 h}{2 π}$

$πr = \frac{h}{mV} = \frac{h}{p} = λ, 3 .14 \times 2 .116 = 6 .64 \overset{O}{A}$

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