An electron in an excited state of Li²⁺ ion has angular momentum 3h/2????. The de Broglie wavelength of the
electron in this state is p????a₀ (where a₀ is the Bohr radius). The value of p is
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Explanation:
Angular momentum, L=
2π
nh
=
2π
3h
⇒n=3
Also, Wavelengthλ=
p
h
=
mvr
hr
=
2π
3h
hr
=
3
2πr
...(1)
But, r=a
0
Z
n
2
....(2)
Substituting (2) in (1), λ=
3
2π
a
0
Z
n
2
=
3
2π
a
0
3
3
2
=2πa
0
⇒p=2 comparing with the given expression pπa
0
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