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The Nikiforov-Uvarov Functional Analysis (NUFA) method recently developed by Ikot et al.52 has been very helpful in providing solutions for exponential type potentials both in relativistic and nonrelativistic wave equations When using this method to solve either the Schrdinger or KleinGordon equation, the energy eigen equation is directly presented in a factorized, closed and compact form. This gives the method an edge over other methods. Meanwhile, the NUFA theory involves solving second order Schrdinger-like differential equation through the analytical combination of Nikiforov-Uvarov (NU) method and functional analysis approach53,54,55. NU is applied to solve a second-order differential equation of the form

$$frac{{d^{2} {uppsi }left( {text{s}} right)}}{{ds^{2} }} + frac{{tilde{tau }left( s right)}}{sigma left( s right)}frac{{d{uppsi }left( {text{s}} right)}}{ds} + frac{{tilde{sigma }left( s right)}}{{sigma^{2} left( s right)}}psi left( s right) = 0$$

(4)

where (sigma (s)) and (widetilde{sigma }left(sright)) are polynomials at most degree two and (widetilde{tau }(s)) is a first-degree polynomial. Tezean and Sever56 latter introduced the parametric form of NU method in the form

$$frac{{d^{2} psi (s)}}{{ds^{2} }} + frac{{alpha_{1} - alpha_{2} s}}{{s(1 - alpha_{3} s)}}frac{{d^{2} psi (s)}}{{ds^{2} }} + frac{1}{{s^{2} (1 - alpha_{3} s)^{2} }}left[ { - U_{1} s^{2} + U_{2} s - U_{3} } right]psi (s) = 0,$$

(5)

where (alpha_{i}) and (xi_{i} (i = 1,2,3)) are all parameters. The differential Eq. (3) has two singularities which is at (s to 0) and (s to frac{1}{{alpha_{3} }}) thus, the wave function can be expressed in the form.

$$Psi_{n} (s) = s^{lambda } (1 - alpha_{3} s)^{v} f(s)$$

(6)

Substituting Eq.(6) into Eq.(5) and simplifying culminate to the following equation,

$$begin{aligned} s(1 - alpha_{3} s)frac{{d^{2} f(s)}}{{ds^{2} }} & + left[ {alpha_{1} + 2lambda - (2lambda alpha_{3} + 2valpha_{3} + alpha_{2} )s} right]frac{df(s)}{{ds}} \ & - alpha_{3} left( {lambda + v + frac{1}{2}left( {frac{{alpha_{2} }}{{alpha_{3} }} - 1} right) + sqrt {frac{1}{4}left( {frac{{alpha_{2} }}{{alpha_{3} }} - 1} right)^{2} + frac{{U_{1} }}{{alpha_{3}^{2} }}} } right)\&quad left( {lambda + v + frac{1}{2}left( {frac{{alpha_{2} }}{{alpha_{3} }} - 1} right) - sqrt {frac{1}{4}left( {frac{{alpha_{2} }}{{alpha_{3} }} - 1} right)^{2} + frac{{U_{1} }}{{alpha_{3}^{2} }}} } right)fleft( s right) \ & + left[ {frac{{lambda (lambda - 1) + alpha_{1} lambda - U_{3} }}{s} + frac{{v(v - 1)alpha_{3} + alpha_{2} v - alpha_{1} alpha_{3} v - frac{{U_{1} }}{{alpha_{3} }} + U_{2} - U_{3} alpha_{3} }}{{left( {1 - alpha_{3} s} right)}}} right]fleft( s right) = 0 \ end{aligned}$$

(7)

Equation(7) can be reduced to a Guassian- hypergeometric equation if and only if the following functions vanished

$$lambda left( {lambda - 1} right) + alpha_{1} lambda - U_{3} = 0$$

(8)

$$upsilon left( {upsilon - 1} right)alpha_{3} + alpha_{2} upsilon - alpha_{1} alpha_{3} upsilon - frac{{U_{1} }}{{alpha_{3} }} + U_{2} - U_{3} alpha_{3} = 0.$$

(9)

Applying the condition of Eq.(8) and Eq.(9) into Eq.(7) results into Eq.(10)

$$begin{aligned} s(1 - alpha_{3} s) & frac{{d^{2} f(s)}}{{ds^{2} }}left[ {alpha_{1} + 2lambda - (2lambda alpha_{3} + 2valpha_{3} + alpha_{2} )s} right]frac{df(s)}{{ds}} \ & ;; - alpha_{3} left( {lambda + v + frac{1}{2}left( {frac{{alpha_{2} }}{{alpha_{3} }} - 1} right) + sqrt {frac{1}{4}left( {frac{{alpha_{2} }}{{alpha_{3} }} - 1} right)^{2} + frac{{U_{1} }}{{alpha_{3}^{2} }}} } right)\ & quadleft( {lambda + v + frac{1}{2}left( {frac{{alpha_{2} }}{{alpha_{3} }} - 1} right) - sqrt {frac{1}{4}left( {frac{{alpha_{2} }}{{alpha_{3} }} - 1} right)^{2} + frac{{U_{1} }}{{alpha_{3}^{2} }}} } right)fleft( s right) = 0 \ end{aligned}$$

(10)

The solutions of Eqs. (8) and (9) are given as

$$lambda = frac{1}{2}left( {left( {1 - alpha_{1} } right) pm sqrt {left( {1 - alpha_{1} } right)^{2} + 4U_{3} } } right)$$

(11)

$$upsilon = frac{1}{{2alpha_{3} }}left( {left( {alpha_{3} + alpha_{1} alpha_{3} - alpha_{2} } right) pm sqrt {left( {alpha_{3} + alpha_{1} alpha_{3} - alpha_{2} } right)^{2} + 4left( {frac{{U_{1} }}{{alpha_{3} }} + alpha_{3} U_{3} - U_{2} } right)} } right)$$

(12)

Equation(10) is the hypergeometric equation type of the form

$$xleft( {1 - x} right)frac{{d^{2} f(s)}}{{ds^{2} }} + left[ {c - left( {a + b + 1} right)x} right]frac{df(x)}{{dx}} - left[ {ab} right]f(x) = 0$$

(13)

where a, b and c are given as follows:

$$a = sqrt {alpha_{3} } left( {lambda + v + frac{1}{2}left( {frac{{alpha_{2} }}{{alpha_{3} }} - 1} right) + sqrt {frac{1}{4}left( {frac{{alpha_{2} }}{{alpha_{3} }} - 1} right)^{2} + frac{{U_{1} }}{{U_{3}^{2} }}} } right)$$

(14)

$$b = sqrt {alpha_{3} } left( {lambda + v + frac{1}{2}left( {frac{{alpha_{2} }}{{alpha_{3} }} - 1} right) - sqrt {frac{1}{4}left( {frac{{alpha_{2} }}{{alpha_{3} }} - 1} right)^{2} + frac{{U_{1} }}{{alpha_{3}^{2} }}} } right)$$

(15)

$$c = alpha_{1} + 2lambda$$

(16)

Setting either a or b equal to a negative integer n, the hypergeometric function f(s) turns to a polynomial of degree n. Hence, the hypergeometric function f(s) approaches finite in the following quantum condition, i.e.,(a = - n) where (n = 0,1,2,3 ldots n_{max }) or (b = - n).

Using the above quantum condition,

$$sqrt {alpha_{3} } left( {lambda + upsilon + frac{1}{2}left( {frac{{alpha_{2} }}{{alpha_{3} }} - 1} right) + sqrt {frac{1}{4}left( {frac{{alpha_{2} }}{{alpha_{3} }} - 1} right)^{2} + frac{{U_{1} }}{{alpha_{3}^{2} }}} } right) = - n$$

(17)

$$lambda + upsilon + frac{1}{2}left( {frac{{alpha_{2} }}{{alpha_{3} }} - 1} right) + frac{n}{{sqrt {alpha_{3} } }} = - sqrt {frac{1}{4}left( {frac{{alpha_{2} }}{{alpha_{3} }} - 1} right)^{2} + frac{{U_{1} }}{{alpha_{3}^{2} }}}$$

(18)

By simplifying Eq.(18), the energy eigen equation using NUFA method is given as

$$lambda^{2} + 2lambda left( {upsilon + frac{1}{2}left( {frac{{alpha_{2} }}{{alpha_{3} }} - 1} right) + frac{n}{{sqrt {alpha_{3} } }}} right) + left( {upsilon + frac{1}{2}left( {frac{{alpha_{2} }}{{alpha_{3} }} - 1} right) + frac{n}{{sqrt {alpha_{3} } }}} right)^{2} - frac{1}{4}left( {frac{{alpha_{2} }}{{alpha_{3} }} - 1} right)^{2} - frac{{U_{1} }}{{alpha_{3}^{2} }} = 0$$

(19)

By substituting Eqs. (9) and (10) into Eq.(6), the corresponding wave equation for the NUFA method as

$$Psi_{n} (s) = N_{n} S^{{frac{{left( {1 - alpha_{1} } right) + sqrt {left( {alpha_{1} - 1} right)^{2} + 4U_{3} } }}{2}}} left( {1 - alpha_{3} } right)^{{frac{{left( {alpha_{3} + alpha_{1} alpha_{3} - alpha_{2} } right) + sqrt {left( {alpha_{3} + alpha_{1} alpha_{3} - alpha_{2} } right)^{2} + 4left( {frac{{U_{1} }}{{alpha_{3}^{2} }} + alpha_{3} U_{3} - U_{2} } right)} }}{{2alpha_{2} }}}} {}_{2}F_{1} (a,b,c;s)$$

(20)

The thermomagnetic energy spectra of 2-Dimensional Schrdinger equation under the influenced of AB and Magnetic field with SPMR potential can be obtained from charged particle Hamiltonian operator of the form

$$left{ {frac{1}{2mu }left( {ihbar nabla - frac{e}{c}vec{A}} right)^{2} + Dleft[ {1 - sigma_{0} left( {frac{{1 + e^{ - alpha r} }}{{1 - e^{ - alpha r} }}} right)} right]^{2} - left( {frac{{c_{1} e^{ - 2alpha r} + c_{2} e^{ - alpha r} }}{{left( {1 - e^{ - alpha r} } right)^{2} }}} right)} right}Rleft( {r, varphi } right) = E_{nm} Rleft( {r, varphi } right)$$

(21)

(E_{nm}) is the thermomagnetic energy spectra, (e) and (mu) represent the charge of the particle and the reduced mass respectively. (c) is the speed of light. Meanwhile, The vector potential (overrightarrow{A}=left({A}_{r},{A}_{phi }, {A}_{z}right)) can be written as the superposition of two terms such that (overrightarrow{A}=overrightarrow{{A}_{1}}+overrightarrow{{A}_{2}}) is the vector potential with azimuthal components such that (overrightarrow{{A}_{1}}=) and (overrightarrow{{A}_{2}}=), corresponding to the extra magnetic flux (Phi_{AB}) generated by a solenoid with (overrightarrow{nabla }.overrightarrow{{A}_{2}}=0) and (overrightarrow{B}) is the magnetic vector field accompanied by (overrightarrow{nabla }times overrightarrow{{A}_{1}}=overrightarrow{B}) ,(overrightarrow{nabla }times overrightarrow{{A}_{2}}=0). The vector potential (overrightarrow{A}) can then be expressed as

$$vec{A} = left( {0,frac{{Be^{ - alpha r} hat{varphi }}}{{1 - e^{ - alpha r} }} + frac{{Phi_{AB} }}{2pi r}hat{varphi },0} right) = left( {frac{{Be^{ - alpha r} hat{varphi }}}{{1 - e^{ - alpha r} }} + frac{{Phi_{AB} }}{2pi r}hat{varphi }} right)$$

(22)

The Laplacian operator and the wave function in cylindrical coordinate is given as

$$begin{aligned} nabla^{2} & = frac{{partial^{2} }}{{partial r^{2} }} + frac{1}{r}frac{partial }{partial r} + frac{1}{{r^{2} }}frac{{partial^{2} }}{{partial varphi^{2} }} + frac{{partial^{2} }}{{partial z^{2} }} \ Psi left( {r,varphi } right) & = frac{1}{{sqrt {2pi r} }}R_{nm} (r)e^{imvarphi } \ end{aligned}$$

(23)

where (m) represents the magnetic quantum number. Substituting Eqs. (23) and (22) into Eq.(21) and with much algebraic simplification gives rise to the Schrdinger -like equation of the form

$$frac{{d^{2} R_{nm} (r)}}{{dr^{2} }} + frac{2mu }{{h^{2} }}left[ begin{gathered} E_{nm} - Dleft[ {1 - sigma_{0} left( {frac{{1 + e^{ - alpha r} }}{{1 - e^{ - alpha r} }}} right)} right]^{2} + left( {frac{{c_{1} e^{ - 2alpha r} + c_{2} e^{ - alpha r} }}{{left( {1 - e^{ - alpha r} } right)^{2} }}} right) - hbar omega_{c} left( {m + xi } right)frac{{e^{ - alpha r} }}{{left( {1 - e^{ - alpha r} } right)r}} hfill \ - left( {frac{{mu omega_{c}^{2} }}{2}} right)frac{{e^{ - 2alpha r} }}{{left( {1 - e^{ - alpha r} } right)^{2} }} - frac{{hbar^{2} }}{2mu }left( {frac{{left( {m + xi } right)^{2} - frac{1}{4}}}{{r^{2} }}} right) hfill \ end{gathered} right]R_{nm} (r) = 0.$$

(24)

where (xi = frac{{Phi_{AB} }}{{phi_{0} }}) is an absolute value containing the flux quantum (phi_{0} = frac{hc}{e}). The cyclotron frequency is represented by (omega_{c} = frac{{evec{B}}}{mu c}). Equation(24) is not exactly solvable due to the presence of centrifugal barrier (frac{1}{{r^{2} }}). In order to provide an analytical approximate solution to Eq.(24), we substitute the modified Greene-Aldrich approximation of the form (frac{1}{{r^{2} }} = frac{{alpha^{2} }}{{left( {1 - e^{ - alpha r} } right)^{2} }}) into Eq.(24) to deal with the centrifugal barrier. Also, using the coordinate transformation (s = e^{ - alpha r}) together with the approximation term, Eq.(24) reduced to the hyper-geometric equation of the form

$$frac{{d^{2} R_{nm} (s)}}{{ds^{2} }} + frac{{left( {1 - s} right)}}{{sleft( {1 - s} right)}}frac{{dR_{nm} (s)}}{ds} + frac{1}{{s^{2} left( {1 - s} right)^{2} }}left{ begin{gathered} - left( {varepsilon^{2} + chi_{1} + 2chi_{1} sigma_{0} + chi_{1} sigma_{0}^{2} - chi_{2} + chi_{5} } right)s^{2} hfill \ + left( {2varepsilon^{2} + 2chi_{1} - 2chi_{1} sigma_{0}^{2} + chi_{3} - chi_{4} } right)s hfill \ - left( {varepsilon^{2} + chi_{1} - 2chi_{1} sigma_{0} + chi_{1} sigma_{0}^{2} + chi_{6} } right) hfill \ end{gathered} right}R_{nm} (s) = 0.$$

(25)

where

$$begin{aligned} varepsilon^{2} & = - frac{{2mu E_{nm} }}{{hbar^{2} alpha^{2} }}begin{array}{*{20}c} , & {chi_{1} = frac{2mu D}{{hbar^{2} alpha^{2} }}} \ end{array} begin{array}{*{20}c} , & {chi_{2} = frac{{2mu c_{1} }}{{hbar^{2} alpha^{2} }}} \ end{array} begin{array}{*{20}c} , & {chi_{3} = frac{{2mu c_{2} }}{{hbar^{2} alpha^{2} }}} \ end{array} \ chi_{4} & = frac{{2mu omega_{c} left( {m + xi } right)}}{hbar alpha }begin{array}{*{20}c} , & {chi_{5} = frac{{mu^{2} omega_{c}^{2} }}{{hbar^{2} alpha^{2} }}} \ end{array} begin{array}{*{20}c} , & {chi_{6} = left( {m + xi } right)^{2} - frac{1}{4}} \ end{array} . \ end{aligned}$$

(26)

Comparing Eq.(25) with NUFA differential equation in Eq.(5), the following polynomial equations can be obtained.

$$begin{gathered} U_{1} = left( {varepsilon^{2} + chi_{1} + 2chi_{1} sigma_{0} + chi_{1} sigma_{0}^{2} - chi_{2} + chi_{5} } right)begin{array}{*{20}c} , & {U_{2} = left( {2varepsilon^{2} + 2chi_{1} - 2chi_{1} sigma_{0}^{2} + chi_{3} - chi_{4} } right)} \ end{array} hfill \ U_{3} = left( {varepsilon^{2} + chi_{1} - 2chi_{1} sigma_{0} + chi_{1} sigma_{0}^{2} + chi_{6} } right),alpha_{1} = alpha_{2} = alpha_{3} = 1. hfill \ end{gathered}$$

(27)

Using equation NUFA in Eqs. (11), (12), (14), (15) and (16) the following polynomial equations can be obtained

$$lambda = sqrt {varepsilon^{2} + chi_{1} - 2chi_{1} sigma_{0} + chi_{1} sigma_{0}^{2} + chi_{6} } ,$$

(28)

$$upsilon = frac{1}{2} + frac{1}{2}sqrt {16chi_{1} sigma_{0}^{2} - 4chi_{2} + 4chi_{5} + 4chi_{6} - 4chi_{3} + 4chi_{4} + 1} ,$$

(29)

$$a = left( begin{gathered} sqrt {varepsilon^{2} + chi_{1} - 2chi_{1} sigma_{0} + chi_{1} sigma_{0}^{2} + chi_{6} } + frac{1}{2} + frac{1}{2}sqrt {16chi_{1} sigma_{0}^{2} - 4chi_{2} + 4chi_{5} + 4chi_{6} - 4chi_{3} + 4chi_{4} + 1} hfill \ + sqrt {varepsilon^{2} + chi_{1} + 2chi_{1} sigma_{0} + chi_{1} sigma_{0}^{2} - chi_{2} + chi_{5} } hfill \ end{gathered} right),$$

(30)

$$b = left( begin{gathered} sqrt {varepsilon^{2} + chi_{1} - 2chi_{1} sigma_{0} + chi_{1} sigma_{0}^{2} + chi_{6} } + frac{1}{2} + frac{1}{2}sqrt {16chi_{1} sigma_{0}^{2} - 4chi_{2} + 4chi_{5} + 4chi_{6} - 4chi_{3} + 4chi_{4} + 1} hfill \ - sqrt {varepsilon^{2} + chi_{1} + 2chi_{1} sigma_{0} + chi_{1} sigma_{0}^{2} - chi_{2} + chi_{5} } hfill \ end{gathered} right),$$

(31)

$$c = left( {1 + 2sqrt {varepsilon^{2} + chi_{1} - 2chi_{1} sigma_{0} + chi_{1} sigma_{0}^{2} + chi_{6} } } right).$$

(32)

using Eq.(19), the thermo-magnetic energy eigen equation

$$begin{aligned} varepsilon^{2} & = frac{1}{4}left{ {frac{{left( {n + frac{1}{2} + frac{1}{2}sqrt {16chi_{1} sigma_{0}^{2} - 4chi_{2} + 4chi_{5} + 4chi_{6} - 4chi_{3} + 4chi_{4} + 1} } right)^{2} + chi_{2} - chi_{5} + chi_{6} - 4chi_{1} sigma_{0} }}{{left( {n + frac{1}{2} + frac{1}{2}sqrt {16chi_{1} sigma_{0}^{2} - 4chi_{2} + 4chi_{5} + 4chi_{6} - 4chi_{3} + 4chi_{4} + 1} } right)}}} right}^{2} \ & ;;; + 2chi_{1} sigma_{0} - chi_{1} - chi_{1} sigma_{0}^{2} - chi_{6} \ end{aligned}$$

(33)

Substituting the parameters of Eq.(26) into Eq.(33), the thermomagnetic energy equation become

$$begin{aligned} E_{nm} & = frac{{h^{2} alpha^{2} }}{2mu }left( {left( {m + xi } right)^{2} - frac{1}{4}} right) + Dleft( {sigma_{0} - 1} right)^{2} \ & ;;; - frac{{h^{2} alpha^{2} }}{8mu }left{ {frac{begin{gathered} left[ {n + frac{1}{2} + frac{1}{2}sqrt {frac{{32mu Dsigma_{0}^{2} }}{{h^{2} alpha^{2} }} - frac{{8mu c_{1} }}{{h^{2} alpha^{2} }} - frac{{8mu c_{2} }}{{h^{2} alpha^{2} }} + frac{{4mu^{2} omega_{c}^{2} }}{{h^{2} alpha^{2} }} + 4left( {m + xi } right)^{2} + frac{{8mu omega_{c} }}{halpha }left( {m + xi } right)} } right]^{2} hfill \ + frac{{2mu c_{1} }}{{h^{2} alpha^{2} }} - frac{{mu^{2} omega_{c}^{2} }}{{h^{2} alpha^{2} }} - frac{{8mu Dsigma_{0} }}{{h^{2} alpha^{2} }} + left( {m + xi } right)^{2} - frac{1}{4} hfill \ end{gathered} }{{left[ {n + frac{1}{2} + frac{1}{2}sqrt {frac{{32mu Dsigma_{0}^{2} }}{{h^{2} alpha^{2} }} - frac{{8mu c_{1} }}{{h^{2} alpha^{2} }} - frac{{8mu c_{2} }}{{h^{2} alpha^{2} }} + frac{{4mu^{2} omega_{c}^{2} }}{{h^{2} alpha^{2} }} + 4left( {m + xi } right)^{2} + frac{{8mu omega_{c} }}{halpha }left( {m + xi } right)} } right]}}} right}^{2} \ end{aligned}$$

(34)

The 2D nonrelativistic energy eigen equation can be obtained with the condition that (omega_{c} = xi = 0), (m = l + frac{1}{2}).

Then Eq.(34) become

$$begin{aligned} E_{nm} & = frac{{h^{2} alpha^{2} lleft( {l + 1} right)}}{2mu } + Dleft( {sigma_{0} - 1} right)^{2} \ & ;;; - frac{{h^{2} alpha^{2} }}{8mu }left{ {frac{{left[ {n + frac{1}{2} + frac{1}{2}sqrt {1 + frac{{32mu Dsigma_{0}^{2} }}{{h^{2} alpha^{2} }} - frac{{8mu c_{1} }}{{h^{2} alpha^{2} }} - frac{{8mu c_{2} }}{{h^{2} alpha^{2} }} + 4lleft( {l + 1} right)} } right]^{2} + frac{{2mu c_{1} }}{{h^{2} alpha^{2} }} - frac{{8mu Dsigma_{0} }}{{h^{2} alpha^{2} }} + lleft( {l + 1} right)}}{{left[ {n + frac{1}{2} + frac{1}{2}sqrt {1 + frac{{32mu Dsigma_{0}^{2} }}{{h^{2} alpha^{2} }} - frac{{8mu c_{1} }}{{h^{2} alpha^{2} }} - frac{{8mu c_{2} }}{{h^{2} alpha^{2} }} + 4lleft( {l + 1} right)} } right]}}} right}^{2} \ end{aligned}$$

(34a)

Substituting (c_{1} = c_{2} = 0).into Eq.(3), then, the potential reduces to Schioberg potential given as

$$Vleft( r right) = Dleft[ {1 - sigma_{0} left( {frac{{1 + e^{ - alpha r} }}{{1 - e^{ - alpha r} }}} right)} right]^{2} .$$

(34b)

Substituting the same condition to Eq.(34) gives the energy-eigen equation for Schioberg potential under the influence of magnetic and AB field as

$$begin{aligned} E_{nm} & = frac{{h^{2} alpha^{2} }}{2mu }left( {left( {m + xi } right)^{2} - frac{1}{4}} right) + Dleft( {sigma_{0} - 1} right)^{2} \ & ;;; - frac{{h^{2} alpha^{2} }}{8mu }left{ {frac{{left[ {n + frac{1}{2} + frac{1}{2}sqrt {frac{{32mu Dsigma_{0}^{2} }}{{h^{2} alpha^{2} }} + frac{{4mu^{2} omega_{c}^{2} }}{{h^{2} alpha^{2} }} + 4left( {m + xi } right)^{2} + frac{{8mu omega_{c} }}{halpha }left( {m + xi } right)} } right]^{2} - frac{{mu^{2} omega_{c}^{2} }}{{h^{2} alpha^{2} }} - frac{{8mu Dsigma_{0} }}{{h^{2} alpha^{2} }} + left( {m + xi } right)^{2} - frac{1}{4}}}{{left[ {n + frac{1}{2} + frac{1}{2}sqrt {frac{{32mu Dsigma_{0}^{2} }}{{h^{2} alpha^{2} }} + frac{{4mu^{2} omega_{c}^{2} }}{{h^{2} alpha^{2} }} + 4left( {m + xi } right)^{2} + frac{{8mu omega_{c} }}{halpha }left( {m + xi } right)} } right]}}} right}^{2} \ end{aligned}$$

(34c)

Substituting (D = 0) into Eq.(3), then the potential reduces to Manning-Rosen potential of the form

$$Vleft( r right) = - left( {frac{{c_{1} e^{ - 2alpha r} + c_{2} e^{ - alpha r} }}{{left( {1 - e^{ - alpha r} } right)^{2} }}} right)$$

(34d)

Substituting the same condition to Eq.(34) gives the energy eigen equation of Manning-Rosen potential under the influence of magnetic and AB fields as

$$begin{aligned} E_{nm} & = frac{{h^{2} alpha^{2} }}{2mu }left( {left( {m + xi } right)^{2} - frac{1}{4}} right) \ & ;;; - frac{{h^{2} alpha^{2} }}{8mu }left{ {frac{{left[ {n + frac{1}{2} + frac{1}{2}sqrt { - frac{{8mu c_{1} }}{{h^{2} alpha^{2} }} - frac{{8mu c_{2} }}{{h^{2} alpha^{2} }} + frac{{4mu^{2} omega_{c}^{2} }}{{h^{2} alpha^{2} }} + 4left( {m + xi } right)^{2} + frac{{8mu omega_{c} }}{halpha }left( {m + xi } right)} } right]^{2} + frac{{2mu c_{1} }}{{h^{2} alpha^{2} }} - frac{{mu^{2} omega_{c}^{2} }}{{h^{2} alpha^{2} }} + left( {m + xi } right)^{2} - frac{1}{4}}}{{left[ {n + frac{1}{2} + frac{1}{2}sqrt { - frac{{8mu c_{1} }}{{h^{2} alpha^{2} }} - frac{{8mu c_{2} }}{{h^{2} alpha^{2} }} + frac{{4mu^{2} omega_{c}^{2} }}{{h^{2} alpha^{2} }} + 4left( {m + xi } right)^{2} + frac{{8mu omega_{c} }}{halpha }left( {m + xi } right)} } right]}}} right}^{2} \ end{aligned}$$

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