| 1. | \(c\sqrt{\dfrac{2m}{E}}\) | 2. | \(\dfrac{1}{c}\sqrt{\dfrac{E}{2m}}\) |
| 3. | \(\sqrt{\dfrac{E}{2m}}\) | 4. | \(c\sqrt{2mE}\) |
| 1. | ![]() |
2. | ![]() |
| 3. | ![]() |
4. | ![]() |
| Statement I: | The de Broglie wavelength associated with a material particle depends on its charge and nature. |
| Statement II: | The wave nature of particles in sub-atomic domain is significant and measurable. |
| 1. | Both Statement I and Statement II are correct. |
| 2. | Both Statement I and Statement II are incorrect. |
| 3. | Statement I is correct but Statement II is incorrect. |
| 4. | Statement I is incorrect but Statement II is correct. |
| 1. | \(\lambda_{\mathrm{e}}>\lambda_{\alpha}\) | 2. | \(\lambda_{\mathrm{e}}=4\lambda_{\alpha}\) |
| 3. | \(\lambda_{\mathrm{e}}=\lambda_{\alpha}\) | 4. | \(\lambda_{\mathrm{e}}<\lambda_{\alpha}\) |
| 1. | 2. | ||
| 3. | 4. |
The de-Broglie wavelength of the thermal electron at \(27^\circ \text{C}\) is \(\lambda.\) When the temperature is increased to \(927^\circ \text{C},\) its de-Broglie wavelength will become:
| 1. | \(2\lambda\) | 2. | \(4\lambda\) |
| 3. | \(\dfrac\lambda2\) | 4. | \(\dfrac\lambda4\) |
An electromagnetic wave of wavelength \(\lambda\) is incident on a photosensitive surface of negligible work function. If '\(m\)' is the mass of photoelectron emitted from the surface and \(\lambda_d\) is the de-Broglie wavelength, then:
| 1. | \( \lambda=\left(\dfrac{2 {mc}}{{h}}\right) \lambda_{{d}}^2 \) | 2. | \( \lambda=\left(\dfrac{2 {h}}{{mc}}\right) \lambda_{{d}}^2 \) |
| 3. | \( \lambda=\left(\dfrac{2 {m}}{{hc}}\right) \lambda_{{d}}^2\) | 4. | \( \lambda_{{d}}=\left(\dfrac{2 {mc}}{{h}}\right) \lambda^2 \) |
An electron is accelerated from rest through a potential difference of \(V\) volt. If the de Broglie wavelength of an electron is \(1.227\times10^{-2}~\text{nm}.\) What will be its potential difference?
| 1. | \(10^{2}~\text{V}\) | 2. | \(10^{3}~\text{V}\) |
| 3. | \(10^{4}~\text{V}\) | 4. | \(10^{5}~\text{V}\) |