Ncert Exercises & Solutions

Question 11.13: What is the
(a) momentum,
(b) speed, and
(c) de Broglie wavelength of an electron with the kinetic energy of 120 eV.

(a)
Hint: p= mv
Step 1: Find the Speed of the electron.

For the electron, we can write the relation for kinetic energy as, $E_{k} = \frac{1}{2} m v^{2}$

Where, v = Speed of the electron

$\begin{array}{l} ∴ v^{2} = \sqrt{\frac{2 e E_{k}}{m}} \\ = \sqrt{\frac{2 \times 1.6 \times 10^{- 19} \times 120}{9.1 \times 10^{- 31}}} \\ = \sqrt{42.198 \times 10^{12}} = 6.496 \times 10^{6} m / s \end{array}$

Step 2: Find the momentum of the electron.
Momentum of the electron, p = mv = 9.1 x $10^{- 31}$ x 6.496 x $10^{6}$
= 5.91 x $10^{- 24}$ kg ms
Therefore, the momentum of the electron is 5.91x $10^{- 24}$ kg m $s^{- 1}$
(b)
Hint: $E_{k}=\frac{1}{2}mv^{2}$
Speed of the electron, v = 6.496 x $10^{6}$ m/s [from part (a)]
(c)
Hint: $\lambda=\frac{h}{p}$
Step: Find de Broglie wavelength of an electron.
De Broglie wavelength of an electron having a momentum P is given as:

$\begin{array}{l} λ = \frac{h}{p} \\ = \frac{6.6 \times 10^{- 14}}{5.91 \times 10^{- 24}} = 1.116 \times 10^{- 10} m \\ = 0.112 n m. \end{array}$

Therefore, the de Broglie wavelength of the electron is 0.112 nm.

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