1.

If the radius of the first Bohr orbit is x, then de- Broglie wavelength of electron in 3rd orbit is nearly

Answer»

3x
9x
`2pi x`
`6pi x`

Solution :`r_(N) = a_(0) (n^(2))/(Z)`. But `mvr_(n) = (NH)/(2pi)`
`:. mv ((a_(0) n^(2))/(Z)) = (nh)/(2pi)`
For 3rd orbit, n = 3 and for hydrogen, Z = 1
`a_(0) = x` (Given, BOHR radius of H-atom)
Hence, `mv(x xx 3^(2))/(1) = (3h)/(2pi) or mv = (h)/(6pi x)`
`LAMDA = (h)/(mv) = (h)/(h//6pi x_ 6pi x`


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