Q. No.An electron has been accelerated by a potential difference of \(V\) volts. The de-Broglie wavelength for the electron is given by:
1. \(\lambda=\dfrac{h}{\sqrt{2 m V e}}\)
2. \(\lambda=\dfrac{h m}{\sqrt{2 V e}}\)
3. \(\lambda=\dfrac{h V}{ \sqrt{2 m e}}\)
4. \(\lambda=\dfrac{hm} { 2 \mathrm{Ve}}\)
1. \(\lambda=\dfrac{h}{\sqrt{2 m V e}}\)
2. \(\lambda=\dfrac{h m}{\sqrt{2 V e}}\)
3. \(\lambda=\dfrac{h V}{ \sqrt{2 m e}}\)
4. \(\lambda=\dfrac{hm} { 2 \mathrm{Ve}}\)
Subtopic: De-broglie Wavelength |
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Which graph correctly shows the variation of momentum \((p)\) of a particle with its associated de-Broglie wavelength \((\lambda)\text{?}\)
| 1. |  |
2. |  |
| 3. |  |
4. |  |
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The total energy of an electron in the \(n^{\text{th}}\) orbit of a hydrogen atom is \((-E_n). \) The de-Broglie wavelength of the electron in this orbit is: (mass of electron = \(m\))
| 1. | \(\dfrac{h}{\sqrt{2mE_n}}\) | 2. | \(\dfrac{h}{\sqrt{mE_n}}\) |
| 3. | \(\dfrac{h}{\sqrt{4mE_n}}\) | 4. | \(\dfrac{hc}{\sqrt{E_n}}\) |
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A proton, an electron an alpha particle have the same energies. Their de-Broglie wavelengths will be compared as :
1. \(\lambda_{\mathrm{p}}>\lambda_{\mathrm{e}}>\lambda_{\mathrm{a}}\)
2. \(\lambda_\alpha<\lambda_{\mathrm{p}}<\lambda_{\mathrm{e}}\)
3. \(\lambda_{\mathrm{p}}<\lambda_{\mathrm{e}}<\lambda_{\mathrm{a}}\)
4. \(\lambda_e>\lambda_\alpha>\lambda_p\)
1. \(\lambda_{\mathrm{p}}>\lambda_{\mathrm{e}}>\lambda_{\mathrm{a}}\)
2. \(\lambda_\alpha<\lambda_{\mathrm{p}}<\lambda_{\mathrm{e}}\)
3. \(\lambda_{\mathrm{p}}<\lambda_{\mathrm{e}}<\lambda_{\mathrm{a}}\)
4. \(\lambda_e>\lambda_\alpha>\lambda_p\)
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If a proton and an electron have the same de-Broglie wavelength, then the approximate ratio of the kinetic energy of the electron to that of the proton is:
1. \(1\)
2. \(1835\)
3. \({1\over 1867}\)
4. \(933.5\)
1. \(1\)
2. \(1835\)
3. \({1\over 1867}\)
4. \(933.5\)
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As the speed of the electron increases, its de-Broglie wavelength:
1. decreases
2. increase
3. remains same
4. either (2) and (3) is possible
1. decreases
2. increase
3. remains same
4. either (2) and (3) is possible
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The equation \(\lambda = {1.227 \over x}~\text{nm}\) can be used to find the de-Broglie wavelength of an electron. In this equation \(x\) stands for:
where,
\(m\)= mass of the electron
\(p\)= momentum of the electron
\(K\)= Kinetic energy of the electron
\(V\) = Accelerating potential in volts for electron
1. \(\sqrt{mK}\)
2. \(\sqrt{P}\)
3. \(\sqrt{K}\)
4. \(\sqrt{V}\)
where,
\(m\)= mass of the electron
\(p\)= momentum of the electron
\(K\)= Kinetic energy of the electron
\(V\) = Accelerating potential in volts for electron
1. \(\sqrt{mK}\)
2. \(\sqrt{P}\)
3. \(\sqrt{K}\)
4. \(\sqrt{V}\)
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In the de-Broglie equation, wavelength \(\mathit{\lambda}\) depends upon mass \(m\) and energy \(E\) according to the relation represented as:
1. \(mE^{1/2}\)
2. \(m^{-1/2}E^{1/2}\)
3. \(m^{-1/2}E^{-1/2}\)
4. \(m^{1/2}E^{-1/2}\)
1. \(mE^{1/2}\)
2. \(m^{-1/2}E^{1/2}\)
3. \(m^{-1/2}E^{-1/2}\)
4. \(m^{1/2}E^{-1/2}\)
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If the kinetic energy of a particle is doubled, the de-Broglie wavelength becomes:
1. 2 times
2. 4 times
3. \(\sqrt{2}\) times
4. \(\left({1/\sqrt{2}}\right)\) times
1. 2 times
2. 4 times
3. \(\sqrt{2}\) times
4. \(\left({1/\sqrt{2}}\right)\) times
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If the kinetic energy of an electron gets tripled, then the de Broglie wavelength associated with it changes by a factor:
1. \(\dfrac{1}{3}\)
2. \(\sqrt{3}\)
3. \(\dfrac{1}{\sqrt{3}}\)
4. \(3\)
1. \(\dfrac{1}{3}\)
2. \(\sqrt{3}\)
3. \(\dfrac{1}{\sqrt{3}}\)
4. \(3\)
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