(A) The de Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{mv}$.Therefore,$\lambda \propto \frac{1}{mv}$.Given: $v_{A} = 3 v_{p}$ and

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(A) The de Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{mv}$.Therefore,$\lambda \propto \frac{1}{mv}$.Given: $v_{A} = 3 v_{p}$ and $\frac{\lambda_{A}}{\lambda_{p}} = \frac{3}{2}$.Using the relation $\frac{\lambda_{A}}{\lambda_{p}} = \frac{m_{p} v_{p}}{m_{A} v_{A}}$,we substitute the given values:$\frac{3}{2} = \frac{m_{p} v_{p}}{m_{A} (3 v_{p})}$.Canceling $v_{p}$ from the numerator and denominator,we get:$\frac{3}{2} = \frac{m_{p}}{3 m_{A}}$.Rearranging for $m_{A}$:$m_{A} = \frac{2}{9} m_{p}$.

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

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The velocity of a particle $A$ is $3$ times the velocity of a proton. If the ratio of the de Broglie wavelengths of the particle $A$ and the proton is $3:2$,the mass of the particle $A$ is (where $m_{p}$ is the mass of the proton).

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A
$\frac{2}{9} m_{p}$
B
$\frac{2}{3} m_{p}$
C
$\frac{2}{5} m_{p}$
D
$\frac{2}{7} m_{p}$

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Dual Nature of Radiation and Matter

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Matter Waves and de Broglie Wavelength

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The velocity of a particle $A$ is $3$ times the velocity of a proton. If the ratio of the de Broglie wavelengths of the particle $A$ and the proton is $3:2$,the mass of the particle $A$ is (where $m_{p}$ is the mass of the proton).

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$A$ material particle with a rest mass $m_0$ is moving with the velocity of light $C$. Then,the wavelength of the de$-$Broglie wave associated with it is

What is the de Broglie wavelength of a nitrogen molecule in air at $300 \;K$ (in $nm$)? Assume that the molecule is moving with the root-mean-square speed of molecules at this temperature. (Atomic mass of nitrogen $= 14.0076 \;u$)

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