11.15 What is the de Broglie wavelength of
(a) Bullet of mass 0.040 kg travelling at the speed of 1.0 km/s,
(b) A ball of mass 0.060 kg moving at a speed of 1.0 m/s, and
(c) A dust particle of mass 1.0 × 10–9 kg drifting with a speed of 2.2 m/s?
11.15 What is the de Broglie wavelength of
(a) Bullet of mass 0.040 kg travelling at the speed of 1.0 km/s,
(b) A ball of mass 0.060 kg moving at a speed of 1.0 m/s, and
(c) A dust particle of mass 1.0 × 10–9 kg drifting with a speed of 2.2 m/s?
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1 Answer
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11.15 Mass of the bullet, m = 0.04 kg
Speed of the bullet, v = 1.0 km/s = 1000 m/s
Planck’s constant, h = 6.626 Js
De Broglie wavelength of the bullet is given by the relation:
= = 1.65 m
Mass of the ball, m = 0.06 kg
Speed of the ball, v = 1.0 m/s
Planck’s constant, h = 6.626 Js
De Broglie wavelength of the bullet is given by the relation:
= = 1.10 m
Mass of the dust particle, m = 1.0 kg
Speed of the dust particle, v = 2.2 m/s
Planck’s constant, h = 6.626 Js
De Broglie wavelength of the bullet is given by the relation:
=
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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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