If E1- E2 and E3 respectively represent the kinetic energies of an electron- an -particle and a proton each having same de-Broglie-s wavelength- then

Kinetic energy E=12mv^2=p^22m Since p=hλ for particle of de Broglie wavelength λ, we can say E=h^22mλ^2 Given, λ_1=λ_2=λ_3 ⇒ E∝1m Since m_α gt;m_p gt;m_e ...

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Standard XII

Physics

Matter Waves

If E1, E2 and...

Question

If $E_{1}, E_{2}$ and $E_{3}$ respectively represent the kinetic energies of an electron, an $α$ -particle and a proton each having same de-Broglie’s wavelength, then

E_{1} > E_{2} > E_{3}

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E_{2} > E_{3} > E_{1}

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E_{1} > E_{3} > E_{2}

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E_{1} = E_{2} = E_{3}

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Solution

The correct option is C $E_{1} > E_{3} > E_{2}$
Kinetic energy $E = \frac{1}{2} m v^{2} = \frac{p^{2}}{2 m}$

Since $p = \frac{h}{λ}$ for particle of de-Broglie wavelength $λ$ , we can say
$E = \frac{h^{2}}{2 m λ^{2}}$

Given, $λ_{1} = λ_{2} = λ_{3}$
$\Rightarrow E \propto \frac{1}{m}$

Since $m_{α} > m_{p} > m_{e}$ i.e $m_{2} > m_{3} > m_{1}$ , we get
$E_{1} > E_{3} > E_{2}$

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If E1,E2 and E3 respectively represent the kinetic energies of an electron, an α-particle and a proton each having same de-Broglie’s wavelength, then

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