If the radius of the 3rd Bohr orbit of hydrogen is r, then the wavelength of an electron in the 6th orbit of a hydrogen atom is equal to
A
6πr
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B
2πr
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C
4πr3
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D
12πr
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Solution
The correct option is C4πr3 According to de-Broglie, circumference of the circular orbit must be an integral number of wavelength.
Thus we can write, 2πrn=nλn ⇒λ6=2πr66 =2π×36r06 ( rn=n2a0z) =2π×(4×9)r0(2×3) =4π×32r03 =4πr3 (r3=9r)