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Relativistic corrections become necessary when the expression for the kinetic energy $\frac{1}{2} m v^2$, becomes comparable with $\mathrm{mc}^2$, where $\mathrm{m}$ is the mass of the particle. At what de-Broglie wavelength, will relativistic corrections become important for an electron?
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Verified Answer
The correct answers are:
$\lambda=10^{-4} \mathrm{~nm}$
,
$\lambda=10^{-6} \mathrm{~nm}$
$\lambda=10^{-4} \mathrm{~nm}$
,
$\lambda=10^{-6} \mathrm{~nm}$
de-Broglie wavelength
As we know that, $\lambda=\frac{\mathrm{h}}{\mathrm{mv}} \Rightarrow \mathrm{v}=\frac{\mathrm{h}}{\mathrm{m} \lambda}$,
$$
\left[\begin{array}{l}
\mathrm{h}=6.6 \times 10^{-34} \mathrm{Js} \\
\mathrm{m}=9 \times 10^{-31} \mathrm{~kg}
\end{array}\right]
$$
Now we consider each option
(a) $\left(\lambda_1=10 \mathrm{~nm}=10 \times 10^{-9} \mathrm{~m}=10^{-8} \mathrm{~m}\right)$
So, $\quad v_1=\frac{6.6 \times 10^{-34}}{\left(9 \times 10^{-31}\right)_{\times 10^{-8}}}$ $=\frac{2.2}{3} \times 10^5$ $\approx 10^5 \mathrm{~m} / \mathrm{s} < \left(3 \times 10^8\right.$ speed of light $)$
(b) $\lambda_2=10^{-1} \mathrm{~nm}=10^{-1} \times 10^{-9} \mathrm{~m}=10^{-10} \mathrm{~m}$
So, $\quad v_2=\frac{6.6 \times 10^{-34}}{\left(9 \times 10^{-31}\right) \times 10^{-10}}$ $\approx 10^7 \mathrm{~m} / \mathrm{s}$
$ < 3 \times 10^8($ speed of ligh $)$
(c) $\lambda_3=10^{-4} \mathrm{~nm}=10^{-4} \times 10^{-9} \mathrm{~m}=10^{-13} \mathrm{~m}$
So, $\quad v_3=\frac{6.6 \times 10^{-34}}{\left(9 \times 10^{-31}\right) \times 10^{-13}}$
$$
\begin{aligned}
&\approx 10^{10} \mathrm{~m} / \mathrm{s} \\
&>3 \times 10^8 \text { (speed of light) }
\end{aligned}
$$
(d) $\lambda_4=10^{-6} \mathrm{~nm}=10^{-6} \times 10^{-9} \mathrm{~m}=10^{-15} \mathrm{~m}$
So,
$$
\begin{aligned}
&\mathrm{v}_4=\frac{6.6 \times 10^{-34}}{9 \times 10^{-31} \times 10^{-15}} \\
&\approx 10^{12} \mathrm{~m} / \mathrm{s} \\
&>3 \times 10^8(\text { speed of ligh })
\end{aligned}
$$
So, options (c) and (d) are correct as $\mathrm{v}_3$ and $\mathrm{v}_4$ is greater than $3 \times 10^8 \mathrm{~m} / \mathrm{s}$ (speed of light).
As we know that, $\lambda=\frac{\mathrm{h}}{\mathrm{mv}} \Rightarrow \mathrm{v}=\frac{\mathrm{h}}{\mathrm{m} \lambda}$,
$$
\left[\begin{array}{l}
\mathrm{h}=6.6 \times 10^{-34} \mathrm{Js} \\
\mathrm{m}=9 \times 10^{-31} \mathrm{~kg}
\end{array}\right]
$$
Now we consider each option
(a) $\left(\lambda_1=10 \mathrm{~nm}=10 \times 10^{-9} \mathrm{~m}=10^{-8} \mathrm{~m}\right)$
So, $\quad v_1=\frac{6.6 \times 10^{-34}}{\left(9 \times 10^{-31}\right)_{\times 10^{-8}}}$ $=\frac{2.2}{3} \times 10^5$ $\approx 10^5 \mathrm{~m} / \mathrm{s} < \left(3 \times 10^8\right.$ speed of light $)$
(b) $\lambda_2=10^{-1} \mathrm{~nm}=10^{-1} \times 10^{-9} \mathrm{~m}=10^{-10} \mathrm{~m}$
So, $\quad v_2=\frac{6.6 \times 10^{-34}}{\left(9 \times 10^{-31}\right) \times 10^{-10}}$ $\approx 10^7 \mathrm{~m} / \mathrm{s}$
$ < 3 \times 10^8($ speed of ligh $)$
(c) $\lambda_3=10^{-4} \mathrm{~nm}=10^{-4} \times 10^{-9} \mathrm{~m}=10^{-13} \mathrm{~m}$
So, $\quad v_3=\frac{6.6 \times 10^{-34}}{\left(9 \times 10^{-31}\right) \times 10^{-13}}$
$$
\begin{aligned}
&\approx 10^{10} \mathrm{~m} / \mathrm{s} \\
&>3 \times 10^8 \text { (speed of light) }
\end{aligned}
$$
(d) $\lambda_4=10^{-6} \mathrm{~nm}=10^{-6} \times 10^{-9} \mathrm{~m}=10^{-15} \mathrm{~m}$
So,
$$
\begin{aligned}
&\mathrm{v}_4=\frac{6.6 \times 10^{-34}}{9 \times 10^{-31} \times 10^{-15}} \\
&\approx 10^{12} \mathrm{~m} / \mathrm{s} \\
&>3 \times 10^8(\text { speed of ligh })
\end{aligned}
$$
So, options (c) and (d) are correct as $\mathrm{v}_3$ and $\mathrm{v}_4$ is greater than $3 \times 10^8 \mathrm{~m} / \mathrm{s}$ (speed of light).
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