(B) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p} = \frac{h}{mv}$.Substituting the given values:$\lambda = \frac{

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(B) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p} = \frac{h}{mv}$.Substituting the given values:$\lambda = \frac{6.63 \times 10^{-34} \ J \cdot s}{(10^{-30} \ kg) \times (2.21 \times 10^6 \ m/s)}$$\lambda = \frac{6.63 \times 10^{-34}}{2.21 \times 10^{-24}} \ m$$\lambda = 3 \times 10^{-10} \ m = 3 \ \mathring{A}$.The wavelength range of $X$-rays is approximately $0.01 \ \mathring{A}$ to $100 \ \mathring{A}$.Since $3 \ \mathring{A}$ falls within the $X$-ray spectrum,the particle will behave like $X$-rays.

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

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$A$ subatomic particle of mass $10^{-30} \ kg$ is moving with a velocity $2.21 \times 10^6 \ m/s$. Under the matter wave consideration,the particle will behave closely like . . . . . . . $(h = 6.63 \times 10^{-34} \ J \cdot s)$

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Dual Nature of Radiation and matter

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Matter Waves and de Broglie Wavelength

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Kinetic energy of a proton is equal to energy $E$ of a photon. Let $\lambda_1$ be the de-Broglie wavelength of the proton and $\lambda_2$ be the wavelength of the photon. If $\frac{\lambda_1}{\lambda_2} \propto E^{n}$,then the value of $n$ is

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The de Broglie wavelength associated with a proton under the influence of an electric potential of $100 \ V$ is

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An electron accelerated through a potential difference $V_1$ has a de-Broglie wavelength of $\lambda$. When the potential is changed to $V_2$, its de-Broglie wavelength increases to $2\lambda$. The value of $(V_1/V_2)$ is equal to

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An electron of mass $m$ and a photon have the same energy $E$. The ratio of the de-Broglie wavelengths associated with them is:

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$A$ subatomic particle of mass $10^{-30} \ kg$ is moving with a velocity $2.21 \times 10^6 \ m/s$. Under the matter wave consideration,the particle will behave closely like . . . . . . . $(h = 6.63 \times 10^{-34} \ J \cdot s)$

Similar Questions

$A$ proton,an electron,and an alpha particle have the same kinetic energies. Their de-Broglie wavelengths will be compared as:

Kinetic energy of a proton is equal to energy $E$ of a photon. Let $\lambda_1$ be the de-Broglie wavelength of the proton and $\lambda_2$ be the wavelength of the photon. If $\frac{\lambda_1}{\lambda_2} \propto E^{n}$,then the value of $n$ is

The de Broglie wavelength associated with a proton under the influence of an electric potential of $100 \ V$ is

An electron accelerated through a potential difference $V_1$ has a de-Broglie wavelength of $\lambda$. When the potential is changed to $V_2$, its de-Broglie wavelength increases to $2\lambda$. The value of $(V_1/V_2)$ is equal to

An electron of mass $m$ and a photon have the same energy $E$. The ratio of the de-Broglie wavelengths associated with them is:

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