The de-Broglie wavelength of an electron accelerated by a voltage of \(50~\text{V}\) is close to:\(\text{(|e|}=1.6\times10^{-19}~\text{C},~{m}_\text{e}=9.1\times10^{-31}~\text{kg, }~{h}=6.6\times10^{-34}~\text{Js})\)
1. \(1.2~\mathring{A}\)
2. \(2.4~\mathring{A}\)
3. \(0.5~\mathring{A}\)
4. \(1.7~\mathring{A}\)
Subtopic:  De-broglie Wavelength |
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The de-Broglie wavelength associated with the electron in the \({n = 4 }\) level is :
1. \(\mathrm{1\over 4}\)th of the de-Broglie wavelength of the electron in the ground state.
2. four times the de-Broglie wavelength of the electron in the ground state
3. two times the de-Broglie wavelength of the electron in the ground state
4. half of the de-Broglie wavelength of the electron in the ground state

 
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A particle \(A\) of mass \(m\) and initial velocity \(v\) collides with a particle \(B\) of mass \(\frac{m}{2}\) which is at rest. The collision is head-on, and elastic. The ratio of the de-Broglie wavelengths \(\lambda_A\) to \(\lambda_B\) after the collision is:
1. \( \frac{\lambda_{\mathrm{A}}}{\lambda_{\mathrm{B}}}=\frac{1}{3} \)
2. \( \frac{\lambda_{\mathrm{A}}}{\lambda_{\mathrm{B}}}=2 \)
3. \( \frac{\lambda_{\mathrm{A}}}{\lambda_{\mathrm{B}}}=\frac{2}{3} \)
4. \( \frac{\lambda_{\mathrm{A}}}{\lambda_{\mathrm{B}}}=\frac{1}{2}\)

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Two electrons are moving with non-relativistic speeds perpendicular to each other. If corresponding de Broglie wavelengths are \(\lambda_1\) and \(\lambda_2,\) their de Broglie wavelength in the frame of reference attached to their center of mass is:
1. \(\frac{1}{\lambda_{com}}=\frac{1}{\lambda_1}+\frac{1}{\lambda_2}\)
2. \(\lambda_{com}=\frac { 2 \lambda_1 \lambda_2 }{ \sqrt {\lambda^2_1 + \lambda^2_2}}\)
3. \(\lambda_{com}= \lambda_1= \lambda_2\)
4. \(\lambda_{com} = \bigg(\frac{\lambda_1 +\lambda_2}{2}\bigg)\)
Subtopic:  De-broglie Wavelength |
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Two particles move at the right angle to each other. Their de-Broglie wavelengths are \(\lambda_1\) and \(\lambda_2\) respectively. The particles suffer perfectly inelastic collision. The de-Broglie wavelength \(\lambda\), of the final particle, is given by:
1. \( \lambda=\frac{\lambda_1+\lambda_2}{2} \)
2. \( \frac{1}{\lambda^2}=\frac{1}{\lambda_1^2}+\frac{1}{\lambda_2^2} \)
3. \( \frac{2}{\lambda}=\frac{1}{\lambda_1}+\frac{1}{\lambda_2} \)
4. \( \lambda=\sqrt{\lambda_1 \lambda_2}\)

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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:
1. \( \lambda_x-\lambda_y \)
2. \( \frac{\lambda_x \lambda_y}{\left|\lambda_x- \lambda_y\right|} \)
3. \( \lambda_x+\lambda_y \)
4. \( \frac{\lambda_x \lambda_y}{\left|\lambda_x+\lambda_y\right|}\)

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Consider an electron in a hydrogen atom revolving in its second excited state (having radius \(4.65~\mathring{\text{A}}\)). The de-Broglie wavelength of this electron is:
1. \(6.6~\mathring{\text{A}}\)
2. \(3.5~\mathring{\text{A}}\)
3. \(9.7~\mathring{\text{A}}\)
4. \(12.9~\mathring{\text{A}}\)

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A particle is moving \(5\) times as fast as an electron. The ratio of the de-Broglie wavelength of the particle to that of the electron is \(1.878 \times 10^{-4}\) . The mass of the particle is close to :
1. \( 4.8 \times 10^{-27} \mathrm{~kg} \)
2. \(1.2 \times 10^{-28} \mathrm{~kg} \)
3. \(9.1 \times 10^{-31} \mathrm{~kg} \)
4. \(9.7 \times 10^{-28} \mathrm{~kg} \)

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Particle \(A\) of mass \(M_A=\frac{m}{2}\) moving along the x-axis with velocity \(v_0\) collides elastically with another particle \(B\) at rest having mass \(M_B=\frac{m}{3}\). If both particles move along the x-axis after the collision, the change \(\Delta \lambda\) in de-Broglie wavelength of particle \(A\), in terms of its de-Broglie wavelength (\(\lambda_0\)) before collision is:
1. \( \Delta \lambda =2 \lambda_0 \)
2. \(\Delta \lambda =\frac{3}{2} \lambda_0 \)
3. \(\Delta \lambda =\frac{5}{2} \lambda_0 \)
4. \(\Delta \lambda =4 \lambda_0\)

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An electron, a doubly ionized helium ion \((\mathrm{He}^{2+})\) and a proton have the same kinetic energy. The correct relationship between their respective de-Broglie wavelengths (\(\lambda_e,\lambda_{\mathrm{He^{2+}}},\) and \(\lambda_p\)) is:

1. \( \lambda_e> \lambda_{\mathrm{H e}^{2+}}>\lambda_p \) 2. \( \lambda_e<\lambda_{\mathrm{H e}^{2+}}=\lambda_p \)
3. \( \lambda_e<\lambda_p<\lambda_{\mathrm{H e}^{2+}} \) 4. \( \lambda_e>\lambda_p>\lambda_{\mathrm{H e}^{2+}} \)
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