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) $(KE)_a = 1\, eV$
(B) $(KE)_b= 0.01\, eV$
(C) $(KE)_c = 4\, eV$
(D) $(KE)_d= 16\, eV$

Choose the correct answer from the options given below:

(B), (A), (C), (D)

The correct answer is Option (2) → (B), (A), (C), (D)

Given: de Broglie wavelength, $\lambda = \frac{h}{\sqrt{2mE}}$

Thus, $\lambda \propto \frac{1}{\sqrt{E}}$

Therefore, greater the kinetic energy, smaller the wavelength.

Given energies:

$E_a = 1\,\text{eV}$, $E_b = 0.01\,\text{eV}$, $E_c = 4\,\text{eV}$, $E_d = 16\,\text{eV}$

Decreasing order of $\lambda$ ⇒ increasing order of $E$:

$E_b < E_a < E_c < E_d$

∴ Order of decreasing de Broglie wavelength:

b > a > c > d