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Question

Chemistry Question on Structure of atom

In hydrogen atom, the de Broglie wavelength of an electron in the second Bohr orbit is : [Given that Bohr radius, a0a_0 = 52.9 pm52.9\, pm ]

A

211.6 pm211.6\, pm

B

211.6 π pm211.6 \,\pi\, pm

C

52.9 π pm52.9 \, \pi\, pm

D

105.8 pm105.8 \,pm

Answer

211.6 π pm211.6 \,\pi\, pm

Explanation

Solution

a=hp mvγ=nh2π P×52.9=2̸×h2̸π P=n(52.9)x a=hh(52.9)π=(52.9π)\begin{array}{l} \qquad a=\frac{h}{p} \\\ m v \gamma=\frac{n h}{2 \pi} \\\ P \times 52.9=\frac{\not 2\times h}{\not {2} \pi} \\\ P=\frac{n}{(52.9) x} \\\ a=\frac{h}{\frac{h}{(52.9)^{\pi}}}=(52.9 \pi) \end{array}