Dual Nature of Radiation and Matter - Online Test
Q1. The work function of a photoelectric material is 3.32 eV. The threshold frequency will be equal to
Answer : Option B
Explaination / Solution:
ϕ0=hν03.32×1.6×10−19=6.6×10−34×ν0ν0=3.32×1.6×10−196.6×10−34=8×1014Hz
Q2.
Light of wavelength 4000 is incident on a metal plate whose work function is 2 eV. The maximum kinetic energy of the emitted photoelectrons would be
Answer : Option B
Explaination / Solution:
ϕ0=2eVλ=4000A∘E=123754000=3.1eVKmax=E−ϕ0=3.1−2=1.1eV
Q3. Given h = 6.6 joule sec, the momentum of each photon in a given radiation is 3.3 kg metre/sec. The frequency of radiation is
Answer : Option C
Explaination / Solution:
λ=hpcν=hpν=pch=3.3×10−29×3×1086.6×10−34=1.5×1013Hz
Q4. In a Milikan’s oil drop apparatus an oil drop of radius 6 m and of density 0.85 kg/m3 is seen to fall freely (without any field). The velocity of drop, given viscosity of air to be 1.83 and neglecting the effect of up thrust force due to air, is
Answer : Option A
Explaination / Solution:
No Explaination.
Q5. If the wavelength of light incident on photo-electric cell be reduced from 4000 to3600,then what will be the change in the cut off potential.
Answer : Option B
Explaination / Solution:
No Explaination.
Q6. If the voltage across the electrodes of a cathode ray tube is 500 volts then energy gained by the electrons is
Answer : Option C
Explaination / Solution:
energy gained by electron
Q7.
If an electron is accelerated by 8.8then electric field required for acceleration is (given specific charge of the electron = 1.76)
Answer : Option B
Explaination / Solution:
a=8.8×1014m/s2em=1.76×1011CKg−1a=Fm=eEm=(em)E8.8×1014=1.76×1011×EE=8.8×10141.76×1011=5000Vm−1=50Vcm−1
Q8. If a beam goes undeflected in Thomson’s experiment, then speed of the electron is (given E = 30 V and
B = 6 x 10-4 T)
Answer : Option C
Explaination / Solution:
If a beam goes undeflected then
Q9.
When an electron enters a magnetic field of 0.01 T with a speed of it describes a circle of radius 6 mm there. Then specific charge of the electron is given by
Answer : Option B
Explaination / Solution:
evB=mv2rem=vrB=1076×10−3×0.01=1.67×1011CKg−1
Q10.
1.If an electron moving with a speed of 2.5 is deflected by an electric field of 1.6 k V perpendicular to its circular path, then e/m for the electron will be (given radius of circlar path = 2.3 m)
Answer : Option C
Explaination / Solution:
Electric field provide required centripetal force for circular motion
