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 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. \(12.2\times 10^{-14}~\text{m}\)