The work function of the photosensitive material is \(4.0~\text{eV}\). The longest wavelength of light that can cause photoelectric emission from the substance is (approximately):
1. \(3100~\text{nm}\)
2. \(966~\text{nm}\)
3. \(31~\text{nm}\)
4. \(310~\text{nm}\)

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 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. \(12.2\times 10^{-14}~\text{m}\)

Light of a frequency of \(1.5\) times the threshold frequency is incident on a photosensitive material. What happens to the photoelectric current when the frequency is cut in half and the intensity is doubled?

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. \(102\times10^{-2}~\text{nm}\)

The number of photons per second on an average emitted by a source of monochromatic light of wavelength \(600~\text{nm}\), when it delivers the power of \(3.3\times 10^{-3}\) watt will be: \((h = 6.6\times10^{-34}~\text{J-s})\)
1. \(10^{16}\)
2. \(10^{15}\)
3. \(10^{18}\)
4. \(10^{17}\)$$

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