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Question

# In hydrogen atom, the de Broglie wavelength of an electron in the second Bohr orbit is [Given that Bohr radius, a0=52.9 pm]:

A
211.6 pm
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B
211.6π pm
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C
52.9π pm
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D
105.8 pm
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Solution

## The correct option is B 211.6π pmBohr radius, a0=52.9 pm n=2, rn=n2a0=(2)2a0=4×52.9 pm=211.6 pm. The angular momentum of an electron in a given stationary state can be expressed as in equation, mvr=nh2π⇒mvr=2h2π=hπ⇒mvrπ=h...(i) We know that, De broglie equation is given as, λ=hmv ⇒λmv=h...(ii) From equation (i) and (ii). We get, λ=πr Putting the value of r, λ=211.6π pm

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