(C) In free space, the de Broglie wavelength is given by $\lambda_0 = h/mv$.When the electron enters the medium, its velocity is reduced by $20\%$.The

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(C) In free space, the de Broglie wavelength is given by $\lambda_0 = h/mv$.When the electron enters the medium, its velocity is reduced by $20\%$.Therefore, the new velocity $v' = v - 0.20v = 0.8v$.The new de Broglie wavelength in the medium is $\lambda = h/mv'$.Substituting $v' = 0.8v$, we get $\lambda = h/(m \times 0.8v) = (1/0.8) \times (h/mv)$.Since $\lambda_0 = h/mv$, we have $\lambda = (1/0.8) \times \lambda_0 = 1.25 \lambda_0$.Comparing this with $\alpha\lambda_0$, we find that $\alpha = 1.25$.

Class 12 Physics Dual Nature of Radiation and Matter Matter Waves and de Broglie Wavelength

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An electron is travelling with a velocity $v$ in free space and when it enters a medium, its velocity is reduced by $20\%$. The de Broglie wavelength of electron in the medium is $\alpha\lambda_0$, where $\lambda_0$ is its de Broglie wavelength in free space. The value of $\alpha$ is . . . . . . .

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A
$1.2$
B
$1$
C
$1.25$
D
$0.75$

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Dual Nature of Radiation and Matter

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Matter Waves and de Broglie Wavelength

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An electron is travelling with a velocity $v$ in free space and when it enters a medium, its velocity is reduced by $20\%$. The de Broglie wavelength of electron in the medium is $\alpha\lambda_0$, where $\lambda_0$ is its de Broglie wavelength in free space. The value of $\alpha$ is . . . . . . .

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An $\alpha$-particle and a proton are accelerated from rest by the same potential. The ratio of their de-Broglie wavelengths is . . . . . .

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$A$ particle $P$ is formed due to a completely inelastic collision of particles $x$ and $y$ having de-Broglie wavelengths $\lambda_x$ and $\lambda_y$ respectively. If $x$ and $y$ were moving in opposite directions,then the de-Broglie wavelength of $P$ is

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