11.31 Crystal diffraction experiments can be performed using X-rays, or electrons accelerated through appropriate voltage. Which probe has greater energy? (For quantitative comparison, take the wavelength of the probe equal to 1 Å, which is of the order of inter-atomic spacing in the lattice) ( = 9.11 × 10–31 kg).
11.31 Crystal diffraction experiments can be performed using X-rays, or electrons accelerated through appropriate voltage. Which probe has greater energy? (For quantitative comparison, take the wavelength of the probe equal to 1 Å, which is of the order of inter-atomic spacing in the lattice) ( = 9.11 × 10–31 kg).
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1 Answer
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11.31 Wavelength of the light emitted from probe, = 1 Å = 1 m
Mass of the electron, = 9.11 kg
Planck’s constant, h = 6.626 Js
Charge of an electron, e = 1.6 C
Speed of light, c = 3
m/sThe kinetic energy of the electron is given as :
= ……………(1)
The de Broglie wavelength is given by
,where p = momentum =
=
Substituting the value of v in equation (1), we get
= = =
=
= ev
= 150.6 eV
Energy of a photon, E’ = = = 12.42 eV= 12.42 keV
Hence, a photon has a greater energy than an electron for the same wavelength.
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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