the EC of an element is 2.6 its velocity is what?.
Answers
Answer:
If the velocity of the electron in Bohr’s first orbit is 2.19 × 106 ms–1, calculate the de Broglie wavelength associated with it.
Answer
According to de Broglie’s equation,
λ = h/mv
Where,
λ = wavelength associated with the electron
h = Planck’s constant
m = mass of electron
v = velocity of electron
Substituting the values in the expression of λ:
λ = 6.626x10-34 / (9.11x10-31)(2.19x106)
=3.32x10-10
= 332 pm
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Using s, p, d notations, describe the orbital with the following quantum numbers.
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Explanation:
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Answer:
The de Broglie wavelength for electron = λ=h/p=c/ν
Given λ
e
=λ
p
KE of electron = 1/2×m×v
2
=1/2×p×v
e
Substituting for p in the above equation, we get
K.E of electron :
2λ
e
hv
e
∴λ
e
=
2×KE
e
h×v
e
The K.E of photon is given by :
λ
p
hc
∴λ
p
=
KE
p
h×c
Given that the wavelengths are equal .
Taking ratio we get :
2×KE
e
v
e
=
KE
p
c
Then,
KE
e
KE
p
=
v
e
2×c
=
1.5×10
8
2×3×10
8