Ncert Exercises & Solutions

Question 11.12: Calculate the
(a) momentum, and
(b) de Broglie wavelength of the electrons accelerated through a potential difference of 56 V.

(a)
Hint: At equilibrium, the kinetic energy of each electron is equal to the accelerating potential.
Step 1: Find the velocity of the electron.
we can write the relation for velocity (v) of each electron as:

$\begin{array}{l} \frac{1}{2} m v^{2} = e V \\ v^{2} = \frac{2 e V}{m} \\ ∴ v = \sqrt{\frac{2 \times 1.6 \times 10^{- 19} \times 56}{9.1 \times 10^{- 31}}} \\ = \sqrt{19.69 \times 10^{12}} = 4.44 \times 10^{6} m / s \end{array}$

Step 2: Find the momentum of the electron.
The momentum of each accelerated electron is given as:
p= mv = 9.1 x 10^-31 x 4.44 x 10⁶ = 4.04 x $10^{- 24} {ms}^{- 1}$

(b)
Hint: $\lambda=\frac{12.27}{\sqrt{V}}$
Step: Find the De Broglie wavelength of an electron.
De Broglie wavelength of an electron accelerating through a potential V is given by the relation:

$\begin{array}{l} λ = \frac{12.27}{\sqrt{V}} A \\ = \frac{12.27}{\sqrt{56}} \times 10^{- 10} m \\ = 0.1639 n m. \end{array}$

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

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