Modern Physics - I
1. In Millikan’s oil drop experiment, an oil drop with a negative charge ‘Q’ in the absence of electric field is found to move down with a terminal velocity ‘v’. When an electric field ‘E’ is applied, the drop is reduced to absolute rest. Now the drop acquires additional negative charge ‘q’ and the drop is found to move up with a terminal velocity of ‘3 v’. The ratio Q : q is
Ans: 1 : 3
SOLUTION:
mg = 6πηrv ............. (1)
mg = EQ ................. (2)
mg + 6πηr (3v) = E (Q + q) ............ (3)
EQ + 3EQ = E (Q + q)
... q = 3Q
... Q : q = 1 : 3
2. A monochromatic light of intensity 6.4 mW emitting 8 × 1015 photon/s falls on a photosensitive metal. If stopping potential is 2 V, then work function is-
Ans: 3 eV
SOLUTION:
3. A silver ball of radius 3 cm is suspended by a thread in a vaccum chamber. UV light of wavelength 310 nm is incident on the ball for some time during which a total energy of 10-11 J falls on the surface. The potential at the centre of the ball due to the release of photo electrons is-
Ans: 0.75 V
SOLUTION:
4. The threshold wavelength for certain metal is 
. When a light of wavelength 
is incident on it, the maximum velocity of photoelectrons is 106 m/s. If the wavelength of the incident radiation is reduced to, 
then the maximum velocity of the photoelectrons in m/s will be
Ans: 2 × 106
SOLUTION:
5. Photons of energies 4.25 eV and 4.7 eV are incident on two metal surfaces A and B respectively. The maximum KE of emitted electrons are respectively TA eV and TB = (TA - 1.5) eV. The ratio de-Broglie wavelengths of photoelectrons from them is λA: λB = 1 : 2, then
a) the work function of A is 3.75 eV b) the work function of A is 2.25 eV
c) the work function of B is 2.73 V d) the work function of B is 4.2 eV
Ans: only b & d are correct
SOLUTION:
6. X-rays of wavelength 0.024 A° are allowed to fall on a metal scatterer. The wavelength of the scattered photon is doubled. The energy of the recoil electron is nearly-
Ans: 2.6 × 105 eV
SOLUTION:
When wavelength is doubled, energy becomes half. The remaining energy is transferred to electron.
7. If 10,000 V applied across an X-ray tube, what will be, the ratio of de- Broglie wavelength of the incident electrons to the shortest wavelength of X-ray produced (e/m of electron is 1.7 × 1011C/Kg)
Ans: 0.1
SOLUTION:
8. A photon of energy 2eV is incident on a metal. The minimum de-Broglie wavelength of the emitted electrons, is λ. When the wavelength of the incident radiation is reduced by 60%, the minimum de-Broglie wavelength of emitted electrons is λ/2. Then the work function of the emitter surface is-
Ans: 1 eV
SOLUTION:. The de-Broglie wavelength of a particle moving with a velocity 2.25 × 108 m/s is equal to the wavelength of a photon. The ratio of kinetic energy of the particle to the energy of photon is (Velocity of light = 3 × 108 m/s)
, where, Z is the atomic number of target and 
υ is the frequency of X-rays, a and b are constants.Ans: 'a' depends on spectral series and member in series but 'b' depends only on spectral series.
SOLUTION:
Where 'n' corresponds to spectral line and 'm' corresponds to spectral series. 'b' is a constant which depends on spectral series. For k series b = 1, l series b = 7
Modern Physics-II
1. In JJ Thomson’s experiment electron beam is undeflected under the action of electric and magnetic fields (crossed fields). Now potential difference between anode and cathode plates is quadrupled and potential difference between condenser plates is tripled keeping distance between two plates constant. The necessary percentage change in the applied magnetic field, for the electron beam to still remain undeflected is
Ans: 50%
Solution:
2. A charged oil drop falls with terminal velocity V0 in the absence of electric field. An electric field of strength E can keep it stationary. If the drop acquires additional charge ‘q’ then it starts moving upwards with velocity V0 / 2, the initial charge on the drop was
Ans: 2 q
Solution:
3. In the Millikan's experiment a charged oil drop moves with the terminal velocities V1 & V2 when the same electric force acts upwards and downwards direction respectively (electric force greater than gravitational force). The terminal velocity in absence of electric field is:
Solution:
4. An energy of 24.6 eV is required to remove one of the electrons from a neutral Helium atom. The energy (in eV) required to remove both the electrons from a neutral Helium atom is
Ans: 79 eV.
Solution:
When one electron is removed, He+ becomes H-like atom so according to Bohr’s model, energy required to remove second electron 13.6 eV × Z2 = 13.6 × 4 = 54.4 eV. So total energy required to remove two electrons is equal to 54.4 + 24.6 = 79 eV.
5. In photo electric experiment, a graph is drawn taking frequency of incident radiation on X - axis and stopping potential on Y - axis.
A) The intercepts on X and Y axes varies from metal to metal
B) The slope of the graph is a constant for all metals.
Ans: A & B are true
Solution:
A) The intercept is equal to threshold frequency, which depends on metal surface.
B) The slope of graph is equal to 
which is always constant.
6. The collector plate in an experiment on photo electric effect is kept vertically above the emitter plate. Light source is above the emitter plate. Light source is put on and saturation photo current is recorded. Now parallel electric & magnetic fields are applied vertically downwards between the plates. Then
Ans: The stopping potential will increase
Solution:
The effect of magnetic field on photo electrons is zero, because magnetic field is parallel to velocity. Electric force which acts upwards, increases velocity of electrons.
7. Electro magnetic radiations described by the equation 
are incident on a photo sensitive surface having work function 2.5 eV. The maximum Kinetic energy of photoelectrons will be
Ans: 1.625 eV
Solution:
8. A source of light is placed above a sphere of radius 10 cm. How many photoelectrons must be emitted by the sphere before emission of photo electrons stop? The energy of incident photon is 4.2 eV and work function of metal is 1.5 eV.
Ans: 1.875 × 108
Solution:
KE = 4.2 - 1.5 = 2.7 v
By losing electrons, sphere becomes more and more positively charged. So when the potential of this charged sphere becomes +2.7 v, then the sphere (cathode) stops emitting electrons
