11.13 What is the
(a) Momentum,
(b) Speed, and
(c) De Broglie wavelength of an electron with kinetic energy of 120 eV.
11.13 What is the
(a) Momentum,
(b) Speed, and
(c) De Broglie wavelength of an electron with kinetic energy of 120 eV.
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1 Answer
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11.13 Kinetic energy of electron, 120 eV = 120 J
Planck’s constant, h = 6.626 Js
Charge of an electron, e = 1.6 C
Mass of electron, m = 9.1 kg
The kinetic energy of electron can be written as m
= = = 6.496 m/s
Momentum of the electron, p = mv = 9.1 6.496 = 5.911 kgm/s
Speed of the electron = 6.496 m/s
De Broglie wavelength of an electron with momentum p is given as
= = 1.121 m = 0.1121 nm
Similar Questions for you
Based on theory
z² × (13.6) (1 - ¼) = 3 × (13.6)
z = 2 . (i)
h/√2mk? = (1/2.3) × h/√2mk?
=> k? = (2.3)²k? = 5.25k? (ii)
Now, k? = E? - Φ
k? = E? - Φ = z²E? - Φ
∴ k? /k? = (10.2 - Φ)/ (4 × 10.2 - Φ) = 1/5.25
=> Φ = 3eV
- (i)
- (ii)
from (i) & (ii)
ev
hu = hu0 + K.E
Cases u = 2u0
h2u0 = hu0 + K.E1
K.E1 = hu0
- (1)
Now, cases 2
h 5u0 = hu0 + k.E2
k.E2 = 4hu0
v2 =
v2 = 2v1
This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:
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