11.12 Calculate the
(a) Momentum, and
(b) De Broglie wavelength of the electrons accelerated through a potential difference of 56 V.
11.12 Calculate the
(a) Momentum, and
(b) De Broglie wavelength of the electrons accelerated through a potential difference of 56 V.
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1 Answer
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11.12 Potential difference, V = 56V
Planck’s constant, h = 6.626 Js
Charge of an electron, e = 1.6 C
Mass of electron, m = 9.1 kg
At equilibrium, the kinetic energy of each electron is equal to the acceleration potential. If v is the velocity of each electron, we can write
=
= = 4.44 m/s
The momentum of each electron = mv = 9.1 4.44 kgm/s = 4.04 kgm/s
De Broglie wavelength of an electron accelerating through a potential V, is given by the relation:
Å = Å = 1.64 Å = 1.64 m = 0.164 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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