if the radius of first bohr orbit of hydrogen atom is x. Then de Broglie wavelength of electron in 3rd orbit is equal to
Answers
Explanation:
De Broglie's wavelength,
λ = the circumference (2πr) of orbit for n=1
That is, an electron in the first bohr orbit.
To find the circumference,
The total energy (E) of an electron in a hydrogen atom,
Energy of an electron,
E = (-) ke² / 2r
For the ground state (n=1)
E = -13.60 eV
E = (13.60 x 1.60^-19)J
E = 2.176^-18 J
Substituting in the total energy formula,
2.176^-18 = ke² / 2r
r = ke² / (2 x 2.176^-18)
r = (9.0^9)(1.60^-19)² / (4.352^-18)
r = 5.29^-11 m
According to De Broglie,
1λ = the circumference of orbit
So,
λ = 2πr = 2 x π x 5.29^-11 m
λ = 3.326^-10 m
The De Broglie wavelength for an electron in the first bohr orbit of a hydrogen atom is 3.326^-10 m
3.5
Answer:
3rd orbit is nearly λ=6πx
Explanation:
According to Bohr's postulate,
mv
n
r
n
=nh/2π (n
th
orbit)
radius of nth orbit, r
n
=n
2
.r
1
given that r
1
=x
thus in third orbit: m.v
3
.r
3
=3h/2π
Debroglie wavelength in 3rd orbit is given by:
λ=h/m
3
v
3
thus λ=2π.r
3
/3
or λ=2π.3
2
x/3
λ=6πx
HOPE THIS HELPS U A LOT