1.

The de Broglie wavelength of electron in second Bohr orbit is exactly equal to

Answer»

the CIRCUMFERENCE of the orbit
Double the circumference of the orbit
Half the circumference of the orbit
thrice the circumference of the orbit

Solution :ANGULAR MOMENTUM `mvr=(nh)/(2PI)`
`mv=(2h)/(2pir)=h/(PIR)`
And `lamda=h/(mv)=(hxxpir)/h=pir`
i.e. half the circumference.


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