The potential energy of a particle of mass m is given by U x - left matrix 0 - - x -1-1 and -2 are the de-Broglie wavelengths of the particle- when 0 - x - 1 and x -1 respectively- If the total energy of particle is 2 E 0- the ratio -1-2 will beA- 1B- 2C- -2D- 1-2

The correct option is C √(2)K.E.=2E_0 E_0 =E_0(for 0≤ x≤ 1)⇒λ _1=h/√(2m E_0) K.E. =2E_0(for x gt;1)⇒λ _2=h/√(4m E_0)⇒λ _1/λ _2=√(2). ...

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Standard XII

Physics

Matter Waves

The potential...

Question

The potential energy of a particle of mass m is given by
$U (x) = {\begin{matrix} E_{0}; & 0 \leq x \leq 1 0; & x > 1 \end{matrix}$

$λ_{1}$ and $λ_{2}$ are the de-Broglie wavelengths of the particle, when $0 \leq x \leq 1$ and $x > 1$ respectively. If the total energy of particle is $2 E_{0}$ , the ratio $\frac{λ_{1}}{λ_{2}}$ will be

2

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1

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\sqrt{2}

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\frac{1}{\sqrt{2}}

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Solution

The correct option is C $\sqrt{2}$
$K . E . = 2 E_{0} - E_{0} = E_{0} (f o r 0 \leq x \leq 1) \Rightarrow λ_{1} = \frac{h}{\sqrt{2 m E_{0}}}$
$K . E . = 2 E_{0} (f o r x > 1) \Rightarrow λ_{2} = \frac{h}{\sqrt{4 m E_{0}}} \Rightarrow \frac{λ_{1}}{λ_{2}} = \sqrt{2}$ .

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Q. The potential energy of a particle of mass m is given by

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The potential energy of a particle of mass m is given by V(x) = ${\begin{matrix} E_{0}, 0 \leq x \leq 1 0, x > 1 \end{matrix}$

$λ_{1}$ and $λ_{2}$ are the de-Broglie wavelengths of the particle, when $0 \leq x \leq 1$ and x > 1respectively. If the total energy of particle is $2 E_{0}$ , then $\frac{λ_{1}}{λ_{2}}$ is

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The potential energy of a particle of mass m is given by U(x)={E0;0≤x≤10;x>1 λ1 and λ2 are the de-Broglie wavelengths of the particle, when 0≤x≤1 and x>1 respectively. If the total energy of particle is 2E0, the ratio λ1λ2 will be

The correct option is C √2K.E.=2E0−E0=E0(for 0≤x≤1)⇒λ1=h√2mE0 K.E.=2E0(for x>1)⇒λ2=h√4mE0⇒λ1λ2=√2.

The correct option is C $\sqrt{2}$
$K . E . = 2 E_{0} - E_{0} = E_{0} (f o r 0 \leq x \leq 1) \Rightarrow λ_{1} = \frac{h}{\sqrt{2 m E_{0}}}$
$K . E . = 2 E_{0} (f o r x > 1) \Rightarrow λ_{2} = \frac{h}{\sqrt{4 m E_{0}}} \Rightarrow \frac{λ_{1}}{λ_{2}} = \sqrt{2}$ .