What did Einstein prove by the photo-electric effect?
1. E = h\(\nu\)
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A light of wavelength \(\lambda \) is incident on the metal surface and the ejected fastest electron has speed \(v.\) If the wavelength is changed to \(\frac{3\lambda}{4},\) then the speed of the fastest emitted electron will be:
1. | \(\sqrt{\frac{4}{3}}v\) | smaller than
2. | \(\sqrt{\frac{4}{3}}\)\(v\) | greater than
3. | \(2v\) |
4. | zero |
The work functions for metals A, B, and C are respectively 1.92 eV, 2.0 eV, and 5 eV. According to Einstein's equation, the metals that will emit photoelectrons for a radiation of wavelength 4100 Å is/are:
1. None
2. A only
3. A and B only
4. All the three metals
The work function of a metal surface is φ = 1.5 eV. If a light of wavelength 5000 Å falls on it, then the maximum K.E. of the ejected electron will be:
1. | 1.2 eV | 2. | 0.98 eV |
3. | 0.45 eV | 4. | 0 eV |
A photosensitive metallic surface has a work function of hν0. If photons of energy 2hν0 fall on this surface, the electrons come out with a maximum velocity of 4 × 106 m/s. When the photon energy is increased to 5hν0, then the maximum velocity of photoelectrons will be:
1. 2 ×107 m/s
2. 2 × 106 m/s
3. 8 × 105 m/s
4. 8 × 106 m/s
According to Einstein's photoelectric equation, the graph between the kinetic energy of photoelectrons ejected and the frequency of incident radiation is:
1. | 2. | ||
3. | 4. |
The current conduction in a discharge tube is due to:
1. electrons only
2. +ve ions and –ve ions
3. –ve ions and electrons
4. +ve ions and electrons
If a light of amplitude A and wavelength λ is incident on a metallic surface, then the saturation current flow is proportional to (assume cut-off wavelength = ):
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Light of wavelength \(3000 ~{\mathring{\mathrm{A}}}\) in Photoelectric effect gives electron of maximum kinetic energy \(0.5 ~\text{eV}\). If the wavelength changes to \(2000 ~{\mathring{\mathrm{A}}}\) then the maximum kinetic energy of emitted electrons will be:
1. | less than \(0.5 ~\text{eV}\) |
2. | \(0.5 ~\text{eV}\) |
3. | greater than \(0.5 ~\text{eV}\) |
4. | the photoelectric effect does not occur |
If the K.E. of an electron and a photon is the same, then the relation between their de-Broglie wavelength will be:
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