A nucleus A, with a finite de-Broglie wavelength , undergoes spontaneous fission into two nuclei B and C of equal mass. B flies in the same direction as that of A, while C flies in the opposite direction with a velocity equal to thalf of that of B. The de-broglie wavelengths of B and C are respectively :-
1.
2.
3.
4.
A particle 'P' is formed due to a completely inelastic collision of particles 'x' and 'y' having de-Broglie wavelengths respectively. If x and y were moving in opposite directions, then the de-Broglie wavelength of 'P' is :-
1.
2.
3.
4.
The electric field of light wave is given as . This light falls on a metal plate of work function 2eV. The stopping potential of the photo-electrons is : Given, E(in eV) =
1. 0.48 V
2. 2.0 V
3. 2.48 V
4. 0.72 V
The magnetic field associated with a light wave is given, at the origin, by . If this light falls on a silver plate having a work function of 4.7 eV, what will be the maximum kinetic energy of the photo electrons ?
1. 7.72 eV
2. 8.52 eV
3. 12.5 eV
4. 6.82 eV
Surface of certain metal is first illuminated with
light of wavelength 1 =350 nm and then, by
light of wavelength 2=540 nm. It is found that
the maximum speed of the photo electrons in
the two cases differ by a factor of 2. The work
function of the metal (in eV) is close to :
(1) 1.8
(2) 1.4
(3) 2.5
(4) 5.6
In a photoelectric effect experiment the threshold wavelength of the light is 380 nm. If the wavelength light is 260 nm, the maximum kinetic energy of emitted electrons will be :
Given in E (in eV) =
1. 1.5 eV
2. 4.5 eV
3. 15.1 eV
4. 3.0 eV
A 2mW laser operated at a wavelength of 500 nm. The number of photons emitted per second is :
1.
2.
3.
4.
A metal plate of area 1 × 10–4 m2 is illuminated
by a radiation of intensity 16 mW/m2.The work
function of the metal is 5eV. The energy of the
incident photons is 10 eV and only 10% of it
produces photo electrons. The number of
emitted photo electrons per second and their
maximum energy, respectively, will be :
[1 eV = 1.6 × 10–19J]
1. and 5 eV
2. and 10 eV
3. and 5 eV
4. and 5 eV
In an electron microscope, the resolution that
can be achieved is of the order of the wavelength
of electrons used. To resolve a width of 7.5 × 10–12m,
the minimum electron energy required is close to :
1. 100 keV
2. 500 keV
3.25 keV
4. 1 keV
If the deBroglie wavelength of an electron is
equal to 10–3 times the wavelength of a photon
of frequency 6 × 1014 Hz, then the speed of
electron is equal to :
(Speed of light = 3 × 108 m/s
Planck's constant = 6.63 × 10–34 J.s
Mass of electron = 9.1 × 10–31 kg)
1. 1.45 × 106 m/s
2. 1.7 × 106 m/s
3. 1.8 × 106 m/s
4. 1.1 × 106 m/s
In a photoelectric experiment, the wavelength
of the light incident on a metal is changed from
300 nm to 400 nm. The decrease in the stopping
potential is close to :
1. 0.5 V
2. 1.0 V
3. 2.0 V
4. 1.5 V
The stopping potential (in volt) as a function of frequency (v) for sodium emitter, is shown in the figure. The work function of sodium, from the data plotted in the figure, will be :
(Given : Planck's constant h = Js, electron charge e =
1. 1.95 eV
2. 1.82 eV
3. 1.66 eV
4. 2.12 eV
When a certain photosensistive surface is illuminated with monochromatic light of frequency v, the stopping potential for the photo current is V0/2. When the surface is illuminated by monochromatic light of frequency 3v, the stopping potential is 2V0. The threshold frequency for photoelectric emission is:
1.
2. v/3
3.
4.
A proton and an -particle (with their masses in the ratio of 1:4 and charges in the ratio of (1:2) are accelerated from rest through a potential difference V. If a uniform magnetic field (B) is set up perpendicular to their velocities, the ratio of the radii of the circular paths described by them will be :
1.
2. 1:2
3. 1:3
4.
An electron (mass m) with an intial velocity is in an electric field . If ,it's de Broglie wavelength at time t is given by
1.
2.
3.
4.