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. \(\frac\lambda2\)
4. \(\frac\lambda4\)

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. \( \lambda=\left(\frac{2 {mc}}{{h}}\right) \lambda_{{d}}^2 \)
2. \( \lambda=\left(\frac{2 {h}}{{mc}}\right) \lambda_{{d}}^2 \)
3. \( \lambda=\left(\frac{2 {m}}{{hc}}\right) \lambda_{{d}}^2\)
4. \( \lambda_{{d}}=\left(\frac{2 {mc}}{{h}}\right) \lambda^2 \)

An electron is accelerated from rest through a potential difference of \(V\) volt. If the de Broglie wavelength of an electron is $\mathrm{}$\(1.227\times10^{-2}~\text{nm}\). what will be its potential difference?
1. \(10^{2}~\text{V}\)
2. \(10^{3}~\text{V}\)
3. \(10^{4}~\text{V}\)
4. \(10^{5}~\text{V}\)

An electron with \(144~\text{eV}\) of kinetic energy has a de-Broglie wavelength that is very similar to?
1. \(102\times10^{-3}~\text{nm}\)
2. \(102\times10^{-4}~\text{nm}\)
3. \(102\times10^{-5}~\text{nm}\)
4. \(102\times10^{-2}~\text{nm}\)

An electron is accelerated through a potential difference of \(10,000~\text{V}\). Its de-Broglie wavelength is, (nearly):\(\left(m_e = 9\times 10^{-31}~\text{kg}\right )\)
1. \(12.2~\text{nm}\)
2. \(12.2\times 10^{-13}~\text{m}\)
3. \(12.2\times 10^{-12}~\text{m}\)
4. \(12.2\times 10^{-14}~\text{m}\)

A proton and an \(\alpha\text{-}\)$$particle are accelerated from rest to the same energy. The de-Broglie wavelength \(\lambda_p\) and \(\lambda_\alpha\)${}_{}$ are in the ratio:
1. \(2:1\)
2. \(1:1\)
3. \(\sqrt{2}:1\)
4. \(4:1\)

An electron of mass m with an initial velocity \(\vec v= v_0\hat i\)$\stackrel{}{}$\( ( v_o > 0 ) \) enters in an electric field\(\vec E = -E_0 \hat i (E_0 = \text{constant}>0)\) at \(t=0\). If \(\lambda_0\)${\mathrm{}}_{}$\(\lambda_0\), is its de-Broglie wavelength initially, then what will be its de-Broglie wavelength at time \(t\)?
1. \(\frac{\lambda_0}{\left(1+ \frac{eE_0}{mv_0}t\right)}\)
2. \(\lambda_0\left(1+ \frac{eE_0}{mv_0}t\right)\)
3. \(\lambda_0 t\)
4. \(\lambda_0\)

The de-Broglie wavelength of a neutron in thermal equilibrium with heavy water at a temperature \(T\) (kelvin) and mass \(m\), is:
1. \(\frac{h}{\sqrt{mkT}}\)
2. \(\frac{h}{\sqrt{3mkT}}\)
3. \(\frac{2h}{\sqrt{3mkT}}\)
4. \(\frac{2h}{\sqrt{mkT}}\)