energy of electronin orbit of H atom is -1.51eV.Wavelength produced by the electron in same orbit if first orbit of H have radius 1)2pie *x 2)3pie*x 3)6pie*x 4)9pie*x
Answers
Answer:
Explanation:
The key to this problem lies with what characterizes a radial node.
Basically, the wave function,
Ψ
(
x
)
, is simply a mathematical function used to describe a quantum object.
The wave function that describes an electron in an atom is actually a product between the radial wave function, which is of interest in your case, and the angular wave function.
The radial wave function depends only on the distance from the nucleus,
r
.
Now, a node occurs when a wave function changes signs, i.e. when its passes through zero. A radial node occurs when a radial wave function passes through zero.
The important thing to remember about nodes is that an electron has zero probability of being located at a node. The probability of an electron being located at a particular point is given by the square of the absolute value of the wave function,
|
Ψ
(
x
)
|
2
.
Since you have zero probability of locating an electron at a node, you can say that you have
|
Ψ
(
x
)
|
2
=
0
→
this is true at nodes
So, you are given the wave function of a 2s-orbital
Ψ
2
s
=
1
2
√
2
π
⋅
√
1
a
0
⋅
(
2
−
r
a
0
)
⋅
e
−
r
2
a
0
and told that at
r
=
r
0
, a radial node is formed. Right from the start, this tells you that you have
|
Ψ
2
s
|
2
=
0
Now, take a look at the wave function again. The only way to get the square of its absolute value equal to zero is if you have
(
2
−
r
a
0
)
=
0
since
Ψ
2
s
=
>
0
1
2
√
2
π
⋅
√
1
a
0
⋅
(
2
−
r
a
0
)
⋅
>
0
e
−
r
2
a
0
This means that you have
2
−
r
a
0
=
0
⇒
r
=
2
⋅
a
0
At
r
=
r
0
, you will have
r
0
=
2
⋅
a
0
Here's how the wave function for the 2s-orbital looks
A 2s-orbital is characterized by the fact that it has no directional properties - you get the exact same value for its wave function regardless of the value of
r
This is why the 2s-orbital is spherical in shape.
Moreover, this tells you that the wave function changes signs at the same distance from the nucleus in all directions, which is why a nodal surface is formed.
The wave function of a 2s-orbital changes signs once, so you only have one nodal surface here.
Answer:Radius of 3rd orbit = r = R(3)^2
= 9 R
[Here R = Radius of first orbit of hydrogen atom]
mvr = nh / (2 π)
h/(mv) = 2πr / n
λ = h / (mv) = 2π × 9R / 3
λ = 6πR
λ = 6 π × (0.529 Å)
λ = 9.97 Å
De-Broglie wavelength is 9.97 Å
Explanation:
Please mark it as THE BRAINLIEST ANSWER!