A particle of mass $1 \times 10^{-30} \mathrm{~kg}$ and electric change $1.6 \times 10^{-19} \mathrm{C}$ has…
A particle of mass $1 \times 10^{-30} \mathrm{~kg}$ and electric change $1.6 \times 10^{-19} \mathrm{C}$ has de-Broglie wavelength $660 \mathrm{~nm}$. Then kinetic energy of this particle is
(Planck's constant, $h=6.6 \times 10^{-34} \mathrm{~J}-\mathrm{s}$ )
$42 \times 10^{-6} \mathrm{eV}$
$2.5 \times 10^{-6} \mathrm{eV}$
$1.3 \times 10^{-6} \mathrm{eV}$
$3.1 \times 10^{-6} \mathrm{eV}$
Solution
de-Broglie wavelength is given by
$
\lambda=h / p
$
$\Rightarrow$ Momentum of particle,
$
\begin{aligned}
& p=\frac{h}{\lambda}=\frac{6.6 \times 10^{-34}}{6.60 \times 10^{-9}}=1 \times 10^{-27} \mathrm{~kg} \mathrm{~m} / \mathrm{s} \\
& \Rightarrow \quad m v=1 \times 10^{-27} \\
& \text { or } \quad v=\frac{1 \times 10^{-27}}{1 \times 10^{-30}} \Rightarrow v=10^3 \mathrm{~m} / \mathrm{s} \\
&
\end{aligned}
$
$\therefore$ Kinetic energy of particle is
$
\begin{aligned}
K & =\frac{1}{2} m v^2 \\
& =\frac{1}{2} \times 1 \times 10^{-30} \times\left(10^3\right)^2 \\
& =0.5 \times 10^{-24} \mathrm{~J}
\end{aligned}
$
As, $\quad 1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}$
Kinetic energy of particle,
$
\begin{aligned}
K & =\frac{0.5 \times 10^{-24}}{1.6 \times 10^{-19}} \mathrm{eV}=0.3125 \times 10^{-5} \\
& =3125 \times 10^{-6} \mathrm{eV} \approx 3.1 \times 10^{-6} \mathrm{eV}
\end{aligned}
$