The potential energy of a particle of mass m is given byUx-E0- 0 - x - 10- 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 2E0- the ratio - 1- 2 will be

K.E.= 2 E_0 E_0 = E_0 (for 0 ≤×≤ 1) ⇒λ _1=h/√(2mE_0) K.E. = 2 E_0 (for x gt; 1) ⇒λ _2=h/√(4mE_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

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\sqrt{2}

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

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Solution

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

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Similar questions

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

V (x) = {\begin{matrix} E_{0} 0 \end{matrix} \begin{matrix} 0 \leq x \leq 1 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}

, find

(λ_{1} / λ_{2})

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

Q. 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

Q. 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

3 E_{0}

, the ratio

\frac{λ_{1}}{λ_{2}}

will be

Q. The potential energy of particle of mass m varies as

U (x) = {\begin{matrix} E_{0} : 0 \leq x \leq 1 0 : x > 1 \end{matrix}

The de Broglie wavelength of the particle in the range

0 \leq x \leq 1 i s λ_{1}

and that in the range

x > 1 i s λ_{2}

. If the total energy of the particle is

2 E_{0}

, find

λ_{1}, / λ_{2} .

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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≤×≤1)⇒λ1=h√2mE0 K.E.=2E0(for x>1)⇒λ2=h√4mE0⇒λ1λ2=√2.

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