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A particle of mass m and energy E, moving in the positive x-direction, encounters a one dimensional potential barrier at x = 0. The barrier is defined by 
V = 0 for x < 0
V = V0 for x > 0 (E > V0)
If the wave function of the particle in the region x < 0 is given as Aeikx + Be-ikx
If B/A = 0.4, then find E/V.
    Correct answer is '1.225'. Can you explain this answer?
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    A particle of mass m and energy E, moving in the positive x-direction,...
    To solve this problem, we need to consider the behavior of the particle as it encounters the potential barrier.

    1. Before the barrier (x < />
    Since the particle is moving in the positive x-direction, it has kinetic energy E. The potential energy V is 0, so the total energy of the particle is E. We can write this as:
    E = KE + V = E + 0 = E

    2. Inside the barrier (0 < x="" />< />
    Inside the barrier, the potential energy V is given by V = -V0, where V0 is a positive constant. The total energy of the particle is still E, so we can write:
    E = KE + V
    E = KE - V0
    KE = E + V0

    3. After the barrier (x > a):
    After the barrier, the potential energy V is 0 again. The total energy of the particle is E, so we can write:
    E = KE + V
    E = KE + 0
    KE = E

    From the above equations, we can see that the kinetic energy of the particle changes as it encounters the barrier. Inside the barrier, the kinetic energy decreases by an amount equal to the potential energy of the barrier (V0).

    This change in kinetic energy can be understood as the particle losing energy to overcome the potential barrier. The amount of energy lost depends on the height of the barrier (V0).

    Note: The above analysis assumes that the particle is a non-relativistic particle. If the particle is a relativistic particle, additional considerations need to be taken into account.
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    A particle of mass m and energy E, moving in the positive x-direction, encounters a one dimensional potential barrier at x = 0. The barrier is defined byV = 0 for x < 0V = V0 for x > 0 (E > V0)If the wave function of the particle in the region x < 0 is given as Aeikx+ Be-ikxIf B/A = 0.4,then find E/V0.Correct answer is '1.225'. Can you explain this answer?
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    A particle of mass m and energy E, moving in the positive x-direction, encounters a one dimensional potential barrier at x = 0. The barrier is defined byV = 0 for x < 0V = V0 for x > 0 (E > V0)If the wave function of the particle in the region x < 0 is given as Aeikx+ Be-ikxIf B/A = 0.4,then find E/V0.Correct answer is '1.225'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A particle of mass m and energy E, moving in the positive x-direction, encounters a one dimensional potential barrier at x = 0. The barrier is defined byV = 0 for x < 0V = V0 for x > 0 (E > V0)If the wave function of the particle in the region x < 0 is given as Aeikx+ Be-ikxIf B/A = 0.4,then find E/V0.Correct answer is '1.225'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle of mass m and energy E, moving in the positive x-direction, encounters a one dimensional potential barrier at x = 0. The barrier is defined byV = 0 for x < 0V = V0 for x > 0 (E > V0)If the wave function of the particle in the region x < 0 is given as Aeikx+ Be-ikxIf B/A = 0.4,then find E/V0.Correct answer is '1.225'. Can you explain this answer?.
    Solutions for A particle of mass m and energy E, moving in the positive x-direction, encounters a one dimensional potential barrier at x = 0. The barrier is defined byV = 0 for x < 0V = V0 for x > 0 (E > V0)If the wave function of the particle in the region x < 0 is given as Aeikx+ Be-ikxIf B/A = 0.4,then find E/V0.Correct answer is '1.225'. Can you explain this answer? in English & in Hindi are available as part of our courses for Physics. Download more important topics, notes, lectures and mock test series for Physics Exam by signing up for free.
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