Question:
A particle of mass \(1\times10^{-30}\ \text{kg}\) and electric charge \(1.6\times10^{-19}\ \text{C}\) has de-Broglie wavelength \(660\ \text{nm}\). Then kinetic energy of this particle is
\[
(\text{Planck's constant, } h=6.6\times10^{-34}\ \text{J s})
\]
A particle of mass \(1\times10^{-30}\ \text{kg}\) and electric charge \(1.6\times10^{-19}\ \text{C}\) has de-Broglie wavelength \(660\ \text{nm}\). Then kinetic energy of this particle is
\[ (\text{Planck's constant, } h=6.6\times10^{-34}\ \text{J s}) \]
\[ (\text{Planck's constant, } h=6.6\times10^{-34}\ \text{J s}) \]
Show Hint
For de-Broglie wavelength problems,
\[
\lambda=\frac{h}{p}
\]
and
\[
K=\frac{p^2}{2m}
=\frac{h^2}{2m\lambda^2}.
\]
This direct relation is very useful in objective questions.
Updated On: Jun 26, 2026
- \(4.2\times10^{-6}\ \text{eV}\)
- \(2.5\times10^{-6}\ \text{eV}\)
- \(1.3\times10^{-6}\ \text{eV}\)
- \(3.1\times10^{-6}\ \text{eV}\)
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Use the de-Broglie wavelength relation.
The de-Broglie wavelength is given by \[ \lambda=\frac{h}{p}. \] Therefore, \[ p=\frac{h}{\lambda}. \] Given, \[ h=6.6\times10^{-34}\ \text{J s}, \] \[ \lambda=660\ \text{nm} =660\times10^{-9}\ \text{m} =6.6\times10^{-7}\ \text{m}. \] Hence, \[ p=\frac{6.6\times10^{-34}}{6.6\times10^{-7}} =10^{-27}\ \text{kg m s}^{-1}. \]
Step 2: Calculate the kinetic energy.
For a non-relativistic particle, \[ K=\frac{p^2}{2m}. \] Substituting \[ p=10^{-27}\ \text{kg m s}^{-1}, \] and \[ m=10^{-30}\ \text{kg}, \] we get \[ K=\frac{(10^{-27})^2}{2\times10^{-30}}. \] \[ K=\frac{10^{-54}}{2\times10^{-30}}. \] \[ K=0.5\times10^{-24}. \] \[ K=5\times10^{-25}\ \text{J}. \]
Step 3: Convert joules into electron-volts.
Since \[ 1\ \text{eV}=1.6\times10^{-19}\ \text{J}, \] \[ K=\frac{5\times10^{-25}}{1.6\times10^{-19}}\ \text{eV}. \] \[ K=\frac{5}{1.6}\times10^{-6}\ \text{eV}. \] \[ K=3.125\times10^{-6}\ \text{eV}. \] \[ K\approx3.1\times10^{-6}\ \text{eV}. \]
Step 4: Final conclusion.
Therefore, the kinetic energy of the particle is \[ \boxed{3.1\times10^{-6}\ \text{eV}} \] Hence, the correct option is \[ \boxed{(4)} \]
The de-Broglie wavelength is given by \[ \lambda=\frac{h}{p}. \] Therefore, \[ p=\frac{h}{\lambda}. \] Given, \[ h=6.6\times10^{-34}\ \text{J s}, \] \[ \lambda=660\ \text{nm} =660\times10^{-9}\ \text{m} =6.6\times10^{-7}\ \text{m}. \] Hence, \[ p=\frac{6.6\times10^{-34}}{6.6\times10^{-7}} =10^{-27}\ \text{kg m s}^{-1}. \]
Step 2: Calculate the kinetic energy.
For a non-relativistic particle, \[ K=\frac{p^2}{2m}. \] Substituting \[ p=10^{-27}\ \text{kg m s}^{-1}, \] and \[ m=10^{-30}\ \text{kg}, \] we get \[ K=\frac{(10^{-27})^2}{2\times10^{-30}}. \] \[ K=\frac{10^{-54}}{2\times10^{-30}}. \] \[ K=0.5\times10^{-24}. \] \[ K=5\times10^{-25}\ \text{J}. \]
Step 3: Convert joules into electron-volts.
Since \[ 1\ \text{eV}=1.6\times10^{-19}\ \text{J}, \] \[ K=\frac{5\times10^{-25}}{1.6\times10^{-19}}\ \text{eV}. \] \[ K=\frac{5}{1.6}\times10^{-6}\ \text{eV}. \] \[ K=3.125\times10^{-6}\ \text{eV}. \] \[ K\approx3.1\times10^{-6}\ \text{eV}. \]
Step 4: Final conclusion.
Therefore, the kinetic energy of the particle is \[ \boxed{3.1\times10^{-6}\ \text{eV}} \] Hence, the correct option is \[ \boxed{(4)} \]
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