The potential energy of a particle of mass m is given by-Ux - 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- then ratio -1-2 will be

Given, Total Energy=2E_0 For, nbsp;(0 ≤ x ≤ 1)→ P.E=E_0 nbsp; K.E = 2E_0 E_0 = E_0 nbsp; ⇒λ_1 = h√(2 mE_0) For, nbsp;(x gt; 1)→ P.E=0 nbsp; ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Introduction to Atomic Structure

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 $2 E_{0}$ , then ratio $\frac{λ_{1}}{λ_{2}}$ will be

2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\sqrt{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $\sqrt{2}$
Given, $Total Energy = 2 E_{0}$

For, $(0 \leq x \leq 1) \to P.E = E_{0}$

$K.E = 2 E_{0} - E_{0} = E_{0}$

$\Rightarrow λ_{1} = \frac{h}{\sqrt{2 m E_{0}}}$

For, $(x > 1) \to P.E = 0$

$K.E = 2 E_{0}$

$\Rightarrow λ_{2} = \frac{h}{\sqrt{4 m E_{0}}}$

$\Rightarrow \frac{λ_{1}}{λ_{2}} = \sqrt{2}$

 Hence, $(C)$ is the correct answer.

Suggest Corrections

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})

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

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

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} .

Join BYJU'S Learning Program

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, then ratio λ1λ2 will be

The correct option is C √2Given, Total Energy=2E0 For, (0≤x≤1)→ P.E=E0 K.E=2E0−E0=E0 ⇒λ1=h√2mE0 For, (x>1)→ P.E=0 K.E=2E0 ⇒λ2=h√4mE0 ⇒λ1λ2=√2  Hence, (C) is the correct answer.