SELECT * FROM question_mgmt as q WHERE id=2786 AND status=1 SELECT id,question_no,question,chapter FROM question_mgmt as q WHERE courseId=3 AND subId=8 AND chapterId=125 and ex_no='1' AND status=1 ORDER BY CAST(question_no AS UNSIGNED)

Question:

An electron and a photon each have a wavelength of 1.00 nm. Find

(a) their momenta,

(b) the energy of the photon, and

(c) the kinetic energy of electron.

Answer:

Wavelength of an electron (λe) and a photon (λp), λe = λp = λ = 1 nm = 1 × 10−9 m

Planck’s constant, h = 6.63 × 10−34 Js

(a) The momentum of an elementary particle is given by de Broglie relation: λ = h/p

∴ p = h/λ

It is clear that momentum depends only on the wavelength of the particle. Since the wavelengths of an electron and a photon are equal, both have an equal momentum.

∴ p = 6.63 x 10-34 / 1 x 10-9 = 6.63 x 10-25 kg m s-1

 

(b) The energy of a photon is given by the relation: E = hc/λ

Where,

Speed of light, c = 3 × 108 m/s

∴ E = 6.63 x 10-34 x 3 x 108 / 1 x 10-9 x 1.6 x 10-19 = 1243.1 eV

Therefore, the energy of the photon is 1.243 keV.

 

(c) The kinetic energy (K) of an electron having momentum p, is given by the relation: K = 1/2 p2 / m

Where,

m = Mass of the electron = 9.1 × 10−31 kg

p = 6.63 × 10−25 kg m s−1

∴ K = 1/2 x (6.63 x 10-25)2 / 9.1 x 10-31 = 2.415 x 10-19 J

= 2.415 x 10-19 / 1.6 x 10-19 = 1.51 eV

Hence, the kinetic energy of the electron is 1.51 eV.


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