The same question has been asked already and was answered very well by Filip Tuomisto. I'll simply copy and paste his answer so all credits to him. (A simple google research would have given you the answer very quickly without having to wait for an answer here btw)
"The concept of effective mass follows from physicists' love for simple relations such as Ohm's law (current density = conductivity x electric field intensity) or Newton's second law of motion (acceleration = force / mass). For a free electron, the mass in the latter is just the electron mass. But, if one wants to write a similar relation for a charge carrier in a crystal lattice, the situation changes. Going through the math (see any solid state physics textbook) allows you to write F = m x a, where the force is charge x electric field, but the mass is no longer the mass of the electron, but reflects the curvature of the conduction band bottom (for electrons) or valence band top (for holes), as it is inversely proportional to the second derivative of the energy as a function of k. This is the so-called effective mass. Again, physicists like simple things, so one often expresses the effective mass as a constant x electron mass, although this has very little (or nothing) to do with the actual physics of the situation. Hence: it is just a practical mathematical construction aimed at simplifying equations (in a similar manner as the reciprocal lattice, for example). The hole itself is also a mathematical construction helping us to avoid using negative values for mass in the simple equations (the effective mass for conduction occurring through the unoccupied electron states in the valence band would be negative, if we didn't invert the charge).
In the majority of cases, the top of the valence band is clearly "flatter" than the bottom of the conduction band. From this follows that the hole effective mass is often larger than the electron effective mass. The top of the valence band tends to be flatter due to the asymmetry of the situation: you are talking about the highest occupied states for electrons. The bottom of the conduction band is formed by the lowest unoccupied states."
The same question has been asked already and was answered very well by Filip Tuomisto. I'll simply copy and paste his answer so all credits to him. (A simple google research would have given you the answer very quickly without having to wait for an answer here btw)
"The concept of effective mass follows from physicists' love for simple relations such as Ohm's law (current density = conductivity x electric field intensity) or Newton's second law of motion (acceleration = force / mass). For a free electron, the mass in the latter is just the electron mass. But, if one wants to write a similar relation for a charge carrier in a crystal lattice, the situation changes. Going through the math (see any solid state physics textbook) allows you to write F = m x a, where the force is charge x electric field, but the mass is no longer the mass of the electron, but reflects the curvature of the conduction band bottom (for electrons) or valence band top (for holes), as it is inversely proportional to the second derivative of the energy as a function of k. This is the so-called effective mass. Again, physicists like simple things, so one often expresses the effective mass as a constant x electron mass, although this has very little (or nothing) to do with the actual physics of the situation. Hence: it is just a practical mathematical construction aimed at simplifying equations (in a similar manner as the reciprocal lattice, for example). The hole itself is also a mathematical construction helping us to avoid using negative values for mass in the simple equations (the effective mass for conduction occurring through the unoccupied electron states in the valence band would be negative, if we didn't invert the charge).
In the majority of cases, the top of the valence band is clearly "flatter" than the bottom of the conduction band. From this follows that the hole effective mass is often larger than the electron effective mass. The top of the valence band tends to be flatter due to the asymmetry of the situation: you are talking about the highest occupied states for electrons. The bottom of the conduction band is formed by the lowest unoccupied states."
The valence band corresponds to more tightly bound 'orbitals' (or atomic bonds) compared to the conduction band, just because it is located deeper in the 'hierarchy' of all the crystal energy bands - both occupied and unoccupied. This means that energy span (=width in eV) of the valence band is lower. The electron momentum span inside the first Brillouin zone is the same for all bands. Qualitatively, a decreased energy span requres an increased mass of the particle - and this can be most easily understood by comparing the dispersion laws for two free particles with different masses. While the dispersion law for electron inside the crystal has little to do with the free electron case, it's all about the width of the energy band.
Also, typically the mobility value is higher for electrons than for holes.
Answer for the same question (Quoted)
The conduction electrons reside in the conduction band and the missing electrons (holes) reside in the valence band of the semiconductor. The conduction band electron effective mass is usually smaller than the valence band hole effective mass.This is one of the reasons that in a semiconductor the electron mobility is usually larger than the hole mobility.
Ref : https://physics.stackexchange.com/questions/376264/why-is-then-the-mobility-of-holes-lesser-than-the-mobility-of-electrons
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Another detailed answer is found @
https://www.tf.uni-kiel.de/matwis/amat/semi_en/kap_2/backbone/r2_3_1.html
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