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 = ${\lambda }_0$):

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

The value of Planck's constant is:

1. $6.63\times {10}^{-34}$ $\mathrm{J}/\mathrm{s}$

2. $6.63\times {10}^{-34}$ $\mathrm{kg}-{\mathrm{m}}^2/\mathrm{s}$

3. $6.63\times {10}^{-34}$ $\mathrm{kg}-{\mathrm{m}}^2$

4. $6.63\times {10}^{-34} J/s$

If particles are moving with the same velocity, then the de-Broglie wavelength is maximum for:
1. proton
2. \(\alpha-\)particle
3. neutron
4. \(\beta-\)particle