Atomic Structure — Medium JEE chemistry MCQ
The de Broglie wavelength of an electron accelerated through a potential difference of $100$ V is approximately (Given: $h = 6.6 \times 10^{-34}$ J·s, $m_e = 9.1 \times 10^{-31}$ kg, $e = 1.6 \times 10^{-19}$ C):
- A. $1.23$ Å
- B. $0.123$ Å
- C. $12.3$ Å
- D. $0.0123$ Å
Solution
Kinetic energy gained by electron: $KE = eV = 1.6 \times 10^{-19} \times 100 = 1.6 \times 10^{-17}$ J
Also, $KE = \frac{1}{2}mv^2$, so $v = \sqrt{\frac{2KE}{m}} = \sqrt{\frac{2 \times 1.6 \times 10^{-17}}{9.1 \times 10^{-31}}}$
$v = \sqrt{3.516 \times 10^{13}} = 5.93 \times 10^6$ m/s
de Broglie wavelength:
$$\lambda = \frac{h}{mv} = \frac{6.6 \times 10^{-34}}{9.1 \times 10^{-31} \times 5.93 \times 10^6}$$
$$\lambda = \frac{6.6 \times 10^{-34}}{5.4 \times 10^{-24}} = 1.22 \times 10^{-10} \text{ m} = 1.22 \text{ Å}$$