SELECT * FROM question_mgmt as q WHERE id=2782 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:

Calculate the

(a) momentum, and

(b) de Broglie wavelength of the electrons accelerated through a potential difference of 56 V.

Answer:

Potential difference, V = 56 V

Planck’s constant, h = 6.6 × 10−34 Js

Mass of an electron, m = 9.1 × 10−31 kg

Charge on an electron, e = 1.6 × 10−19 C

(a) At equilibrium, the kinetic energy of each electron is equal to the accelerating potential, i.e., we can write the relation for velocity (v) of each electron as:

1/2 mv2 = eV

v2 = 2eV / m

The momentum of each accelerated electron is given as:

p = mv

= 9.1 × 10−31 × 4.44 × 106

= 4.04 × 10−24 kg m s−1

Therefore, the momentum of each electron is 4.04 × 10−24 kg m s−1 .

 

(b) De Broglie wavelength of an electron accelerating through a potential V, is given by the relation:

Therefore, the de Broglie wavelength of each electron is 0.1639 nm.


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