Question

A particle of mass $1 \times 10^{-30}$ kg and electric charge $1.6 \times 10^{-19}$ C has de-Broglie wavelength 660 nm. Then kinetic energy of this particle is (Planck's constant, h = $6.6 \times 10^{-34}$ J-s)

• A. $4.2 \times 10^{-6}$ eV
• B. $2.5 \times 10^{-6}$ eV
• C. $1.3 \times 10^{-6}$ eV
• D. $3.1 \times 10^{-6}$ eV

Hint:

de-Broglie wavelength: $\lambda = h/p$.
Kinetic energy: KE $= p^2/(2m)$.
Combining these: KE $= h^2 / (2m\lambda^2)$.
Ensure units are consistent (SI units for $h, m, \lambda$ will give KE in Joules).
Conversion: $1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}$.
The electric charge given is likely for the J to eV conversion factor.

Answer 1

The Correct Option is D

The de-Broglie wavelength ($\lambda$) of a particle is related to its momentum ($p$) by: $\lambda = \frac{h}{p}$, where $h$ is Planck's constant. So, $p = \frac{h}{\lambda}$. The kinetic energy (KE) of a non-relativistic particle is related to its momentum by: KE $= \frac{p^2}{2m}$, where $m$ is the mass of the particle. Substituting $p = h/\lambda$: KE $= \frac{(h/\lambda)^2}{2m} = \frac{h^2}{2m\lambda^2}$. Given values: Mass $m = 1 \times 10^{-30} \text{ kg}$. de-Broglie wavelength $\lambda = 660 \text{ nm} = 660 \times 10^{-9} \text{ m} = 6.6 \times 10^{-7} \text{ m}$. Planck's constant $h = 6.6 \times 10^{-34} \text{ J-s}$. Electric charge $q = 1.6 \times 10^{-19} \text{ C}$ (this is the elementary charge, $e$. It will be used for converting Joules to eV). Calculate KE in Joules first: KE $= \frac{(6.6 \times 10^{-34} \text{ J-s})^2}{2 \times (1 \times 10^{-30} \text{ kg}) \times (6.6 \times 10^{-7} \text{ m})^2}$. KE $= \frac{(6.6)^2 \times (10^{-34})^2}{2 \times 10^{-30} \times (6.6)^2 \times (10^{-7})^2}$. KE $= \frac{(6.6)^2 \times 10^{-68}}{2 \times 10^{-30} \times (6.6)^2 \times 10^{-14}}$. The $(6.6)^2$ terms cancel out: KE $= \frac{10^{-68}}{2 \times 10^{-30} \times 10^{-14}} = \frac{10^{-68}}{2 \times 10^{-44}}$. KE $= \frac{1}{2} \times 10^{-68 - (-44)} = \frac{1}{2} \times 10^{-68 + 44} = \frac{1}{2} \times 10^{-24} \text{ J}$. KE $= 0.5 \times 10^{-24} \text{ J}$. Now, convert KE from Joules to electron-volts (eV). $1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}$. So, KE (in eV) $= \frac{\text{KE (in J)}}{1.6 \times 10^{-19}}$. KE (in eV) $= \frac{0.5 \times 10^{-24}}{1.6 \times 10^{-19}}$. KE (in eV) $= \frac{0.5}{1.6} \times 10^{-24 - (-19)} = \frac{0.5}{1.6} \times 10^{-24 + 19} = \frac{0.5}{1.6} \times 10^{-5}$. $\frac{0.5}{1.6} = \frac{5}{16}$. $\frac{5}{16} = 0.3125$. So, KE $= 0.3125 \times 10^{-5} \text{ eV} = 3.125 \times 10^{-6} \text{ eV}$. This value is approximately $3.1 \times 10^{-6}$ eV. This matches option (d). \[ \boxed{3.1 \times 10^{-6} \text{ eV}} \]