Question

Assuming Heisenberg Uncertainty principle to be true what could be the minimum uncertainty in de-Broglie wavelength of a moving electron accelerated by potential difference of 6 V whose uncertainty in position is $\dfrac{7}{{22}}{\text{nm}}{\text{.}}$
A.$6.25\mathop {\text{A}}\limits^ \circ$
B.$6\mathop {\text{A}}\limits^ \circ$
C.$0.625\mathop {\text{A}}\limits^ \circ$
D.$0.3125\mathop {\text{A}}\limits^ \circ$

Answer 1

Heisenberg uncertainty principle states that position and momentum cannot be determined simultaneously. The formula for the Heisenberg uncertainty principle and the relationship between de Broglie wavelength and momentum should be used. Substituting the values, will get the uncertainty in wavelength.
Formula used: ${{\lambda = }}\dfrac{{\text{h}}}{{\text{p}}}$ and $\Delta {\text{p}} \times \Delta {\text{x}} \geqslant \dfrac{{\text{h}}}{{4\pi }}$
${\text{E = }}\dfrac{{{{\text{p}}^{\text{2}}}}}{{{\text{2m}}}}$ and ${\text{E = eV}}$
where, $\lambda$ is de Broglie wavelength, h is planck's constant, p is momentum, $\Delta {\text{p}}$ is change in momentum and $\Delta {\text{x}}$ is change in position.
m is mass of electron that is $9.1 \times {10^{ - 31}}{\text{Kg}}$, e is charge on electron that is $1.67 \times {10^{ - 19}}{\text{C}}$.

Complete step by step answer:
We have been given uncertainty in position and we have to calculate the uncertainty in wavelength but Heisenberg has given us the relation between uncertainty in position and momentum. So we have to derive a relation for the same using de Broglie wavelength:
${{\lambda = }}\dfrac{{\text{h}}}{{\text{p}}}$
Differentiating both sides will give us,
$\Delta \lambda = h\dfrac{{\Delta p}}{{{p^2}}}$
Because differentiation of $\dfrac{1}{p}{\text{ is - }}\dfrac{{\Delta p}}{{{p^2}}}$.
We can ignore negative sign here. Using Heisenberg formula and rearranging we will get, $\Delta p \geqslant \dfrac{h}{{4\pi \times \Delta x}}$
It is given that electrons are accelerated with a potential of 6V. Using energy momentum formula:
${\text{E = }}\dfrac{{{{\text{p}}^{\text{2}}}}}{{{\text{2m}}}}$ and ${\text{E = eV}}$ electrical energy is equal to charge into potential. Rearranging, we will get the final equation as,
${{\text{p}}^{\text{2}}}{\text{ = 2meV}}$
We will substitute the values in S.I units in the wavelength equation and we will get:
$\Delta \lambda = \dfrac{{{{(6.626 \times {{10}^{ - 34}})}^2} \times 22}}{{4\pi 2(9.1 \times {{10}^{ - 31}})(1.6 \times {{10}^{ - 19}})(6)(7 \times {{10}^{ - 9}})}}$
Solving this, we get:

= 0.625 \times {10^{ - 10}}m \\\ = 0.625\mathop {\text{A}}\limits^ \circ \\\ \end{gathered} $$ The unit conversion we have used is, $$1{\text{ m = 1}}{{\text{0}}^{10}}\mathop {\text{A}}\limits^ \circ = {10^9}nm$$ **Hence the correct option is C.** **Note:** The uncertainty relation can be derived for other variables as well. As in the question we have derived for uncertainty in wavelength. We can even derive the relation likewise for calculation of uncertainty in energy as well.