Kinetic energies of four particles a, b, c, d of equal masses are given below. Arrange them in order of their decreasing de-Broglie wavelengths.
(A) \( (KE)_a = 1\,\mathrm{eV} \)
(B) \( (KE)_b = 0.01\,\mathrm{eV} \)
(C) \( (KE)_c = 4\,\mathrm{eV} \)
(D) \( (KE)_d = 16\,\mathrm{eV} \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (B), (A), (C), (D)
The de-Broglie wavelength of a particle of mass \( m \) and kinetic energy \( K \) is
\( \displaystyle \lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}} \;\Rightarrow\; \lambda \propto \frac{1}{\sqrt{K}} \quad (m \text{ equal}) \)
So the larger the kinetic energy, the shorter the wavelength. Taking \( \lambda \propto 1/\sqrt{K} \) with \( K \) in eV:
\( \displaystyle (A)\ (KE)_a = 1\,\mathrm{eV}:\ \frac{1}{\sqrt{1}} = 1 \qquad (B)\ (KE)_b = 0.01\,\mathrm{eV}:\ \frac{1}{\sqrt{0.01}} = 10 \)
\( \displaystyle (C)\ (KE)_c = 4\,\mathrm{eV}:\ \frac{1}{\sqrt{4}} = 0.5 \qquad (D)\ (KE)_d = 16\,\mathrm{eV}:\ \frac{1}{\sqrt{16}} = 0.25 \)
In decreasing order of wavelength, \( \lambda_B \gt \lambda_A \gt \lambda_C \gt \lambda_D \), that is (B), (A), (C), (D).