Answer
Verified
402k+ views
Hint: 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:
\[\begin{gathered}
= 0.625 \times {10^{ - 10}}m \\
= 0.625\mathop {\text{A}}\limits^ \circ \\
\end{gathered} \]
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.
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:
\[\begin{gathered}
= 0.625 \times {10^{ - 10}}m \\
= 0.625\mathop {\text{A}}\limits^ \circ \\
\end{gathered} \]
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.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
Give 10 examples for herbs , shrubs , climbers , creepers
Select the word that is correctly spelled a Twelveth class 10 english CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Fill the blanks with proper collective nouns 1 A of class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE
Change the following sentences into negative and interrogative class 10 english CBSE