The movement of holes means the movement of electrons in opposite direction, so in Hall Effect the particles on which force is applied finally must be electrons, so there must be no positive charge appearing in this scenario.
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2$\begingroup$ The model that works uses holes. $\endgroup$– John DotyCommented Feb 6 at 19:24
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3$\begingroup$ The "electrons" in condensed matter don't generally behave like electrons in a vacuum. $\endgroup$– John DotyCommented Feb 6 at 19:39
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1$\begingroup$ Semiconductors, doped or not, are electrically neutral. And holes are a very necessary part of understanding semiconductors and even some metals. E.g. under certain conditions in aluminum the hall effect will indicate a positive charge carrier. Think on that for a bit... $\endgroup$– Jon CusterCommented Feb 6 at 19:48
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1$\begingroup$ physics.stackexchange.com/questions/538629/… may be of some use as well. In solid state physics you really need to use holes. $\endgroup$– Jon CusterCommented Feb 6 at 20:01
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2$\begingroup$ @SolomonSlow - without treating holes as an equal to electrons you have difficulty using detailed balance to discuss majority and minority carrier concentrations in a doped semiconductor in the first place. And certainly if you can't get there in a bulk region you aren't getting to a junction either. $\endgroup$– Jon CusterCommented Feb 6 at 20:13
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Electrons at the top of the valence band have negative effective mass. Holes are just a way of inverting the band's energy diagram, speaking in terms of the positive charges of the lack of electrons instead of the negative charges of the electrons themselves.
If you want to understand the Hall effect in terms of electrons instead, take the negative effective mass of an electron, put it into the second Newton's law, and see where the acceleration will be directed with respect to the electric field.
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$\begingroup$ the electrons are getting deflected in the same direction as holes. i am not getting the right hall voltage if the electrons are getting deflected in same direction as holes $\endgroup$ Commented Feb 6 at 19:36
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3$\begingroup$ @RaqibSyed Did you change all the relevant signs? Mass, charge, drift velocity. $\endgroup$– RuslanCommented Feb 6 at 19:39
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2$\begingroup$ Remember $F = ma$, so if the magnetic force is the same yet $m$ becomes $-m$, what happens to $a$? $\endgroup$ Commented Feb 6 at 19:49
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1$\begingroup$ @RaqibSyed see the link to the article about effective mass that I gave in the body of my answer. $\endgroup$– RuslanCommented Feb 6 at 19:51
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2$\begingroup$ @RaqibSyed no, the band structure itself is a product of interference of the electron waves scattered by the crystal lattice. The effective mass is introduced in terms of this band structure. In particular, the transition of an electron from the bottom of a band to the top (changing its effective mass from positive to negative) can give rise to Bloch oscillation, which is a completely non-classical behavior. $\endgroup$– RuslanCommented Feb 6 at 19:59