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.	smaller than \(\sqrt{\frac{4}{3}}v\)
2.	greater than \(\sqrt{\frac{4}{3}}\)\(v\)
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:

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 = ${\lambda }_0$):

1. $A^2,$ $if$ $\lambda$ $>$ ${\lambda }_0$

2. $A^2,$ $if$ $\lambda$ $<$ ${\lambda }_0$

3. $A,$ $if$ $\lambda$ $>$ ${\lambda }_0$

4. $A,$ $if$ $\lambda$ $<$ ${\lambda }_0$

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:

If the K.E. of an electron and a photon is the same, then the relation between their de-Broglie wavelength will be:

1. ${\lambda }_{ph}$ $<$ ${\lambda }_e$

2. ${\lambda }_{ph}$ $=$ ${\lambda }_e$

3. ${\lambda }_{ph}$ $>$ ${\lambda }_e$

4. ${\lambda }_{ph}$ $=2$ ${\lambda }_e$

The total energy of an electron is \(3.555~\text{MeV}\). Its kinetic energy will be:
1.  \(3.545~\text{MeV}\)
2.  \(3.045~\text{MeV}\)
3.  \(3.5~\text{MeV}\)
4.  none of the above