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As a model potential function, one can take the potential in w

Question

As a model potential function, one can take the potential in which an alpha particle moves inside a large nucleus as $V (x) =$ $(K_{1} x^{2} - V_{0}) e^{- k_{2} x^{2}}$ (where $K_{1,} K_{2}$ are positive constants). It has

only one extreme viz

x = 0

two extreme viz

x =

\pm \frac{K _{1} - K _{2} V _{0}}{K _{2}}

three extreme viz

x = 0

and

x =

\pm \frac{K _{1} + K _{2} V _{0}}{K _{1} K _{2}}

four extreme viz

x = 0

x =

\pm \frac{K _{1} + K _{2} V _{0}}{K _{1} K _{2}}

and x =

\pm \frac{K _{2} - K _{1} V _{0}}{K _{1}^{2} K _{2}^{2}}

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As a model potential function, one can take the potential in which an alpha particle moves inside a large nucleus as

V (x) =

(K_{1} x^{2} - V_{0}) e^{- k_{2} x^{2}}

(where

K_{1,} K_{2}

are positive constants). It has

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Text solutionVerified

Given,

V (x) = (K_{1} x^{2} - V_{0}) e^{- K_{2} x^{2}}

F = - \frac{d V}{d x} = 2 K_{1} x e^{- K_{2} x^{2}} + (- 2 K_{2} x) e^{- K_{2} x^{2}} (K_{1} x^{2} - V_{0}) = 0

\Rightarrow x = 0

2 K_{1} - 2 K_{2} (K_{1} x^{2} - V_{0}) = 0

\Rightarrow K_{1} K_{2} x^{2} = K_{2} V_{0} + K_{1}

x = \pm \frac{K _{1} + K _{2} V _{0}}{K _{1} K _{2}}

Hence,

x = 0 or \pm \frac{K _{1} + K _{2} V _{0}}{K _{1} K _{2}}

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Question Text	As a model potential function, one can take the potential in which an alpha particle moves inside a large nucleus as $V (x) =$ $(K_{1} x^{2} - V_{0}) e^{- k_{2} x^{2}}$ (where $K_{1,} K_{2}$ are positive constants). It has
Answer Type	Text solution:1
Upvotes	150