WorksheetsDual Nature of matter & Radiation, Atoms and nuclei, Electronic
Total questions: 40
Worksheet time: 21mins
The photoelectric work function for a metal surface is 4.14eV.The cutoff wavelength for this
4125 A0
2062.5 A0
3000 A0
6000 A0
Maximum K.E of emitted electron depends on the frequency of incident photon when frequency of incident photons is
equal to threshold frequency
half of threshold frequency
greater than threshold frequency
one third of threshold frequency
The slope of frequency of incident ray and stopping potential for a given surface
h
h/e
eh
e
The photoelectric cut off voltage in a certain experiment is 1.5V. What is the max. K.E of photoelectrons emitted?
1.5 eV
0.5 eV
2 eV
1eV
THE SLOPE OF the cut off voltage versus frequency of incident light is found to be 4.12 × 10−15 Vs.Given e =1.6 × 10−19 C CALCULATE THE VALUE 0f plancks constant
5.592 × 10−34 Js
6.592 × 10−34 Js
5.592
6.592
What is the momentum of an electron with K.E of 120eV
5.88 × 10−24 kg m/s
6.88 × 10−24 kg m/s
120 kg m/s
1.6 × 102 kg m/s
stopping potential depends on intensity of incident light
true
photons
frequency
none
Kinetic energy of emitted electrons depends upon :
(a) frequency
(b) intensity
(c) nature of atmosphere surrounding the electrons
(d) none of these
De-Broglie wavelength of a body of mass m and kinetic energy E is given by (symbols have their usual meanings):
(a) h/√2mE
(b) h/2mE
(c)√ 2mE/h
(d) h/mE
The work function of photoelectric material is 3.3 eV. The threshold frequency will be equal to:
(a) 8 × 1014 Hz
(b) 8 × 1010 Hz
(c) 5 × 1010 Hz
(d) 4 × 1014 Hz
The momentum of an electron that emits a wavelength of 2 Å. will be:
(a) 6.4 × 10-36 kgms-1
(b) 3.3 × 10-24 kgms-1
(c) 3.3 × 10-34 kgms-1
(d) none of these
Name the series of hydrogen spectrum which lie in the ultraviolet and visible region
(a)
The ground state energy of hydrogen atom is -13.6 eV. What are the kinetic and
potential energies of the electron in this state
13.6 eV and 27.2 eV
-13.6eV and 27.2 eV
13.6 and -27.2 eV
None
The total energy of an electron in the first excited state of hydrogen atom is about - 3.4 eV. Its kinetic energy in this state is:
3.4 eV
-6.8 eV
-3.4 eV
6.8 eV
What do these isotopes of carbon all have in common?
126C, 136C, 146C
they all have the same number of neutrons & mass number
they all have the same atomic number and neutrons
they all have the same atomic number and electrons
they all have the same number of protons, atomic number, and mass number
The energy equivalent in joules of the mass of
2.0 *10-31 𝑘𝑔 _____.
2.0 *10-31𝐽
6.0 *10-23 𝐽
1.8 * 10-14𝐽
8.2 * 10-14 𝐽
is an example of
nuclear fission
radioactive decay
nuclear fusion
artificial transmutation
If the binding energy of the deutrium is 2.23 MeV . The mass defect given in a.m.u. is
-0.0024
-0.0012
0.0012
0.0024
r1 and r2 are the radii of atomic nuclei of mass numbers 64 and 27 respectively. The ratio (r1 /r2) is
64/27
27/64
4/3
1
What is the name of the circuit as shown?
Half-wave rectifier
Full-wave rectifier
Full-wave bridge rectifier
Secondary rectifier
Doping of Semiconductors
materials which can conduct electricity better than insulator, but not as well as conductors.
The process of adding of a certain amount of specific impurities atoms in the pure semiconductor.
when 4 electrons in outermost orbit.
which statement is correct? (the dark circles represent the electrons)
p-type of the diode is connected to the negative terminal.
Electron moves across the p-n junction, makes the diode forward biased
current will not flow in the circuit
reverse biased circuit
Let np and ne be the number density of holes and conduction electrons respectively in a semiconductor. Then,
np > ne in an intrinsic semiconductor, I < Ip + Ie
np = ne in an extrinsic semiconductor, I > Ip + Ie
np = ne in an intrinsic semiconductor, I = Ip + Ie
np > ne in an intrinsic semiconductor, I = 0( Here, Ip = current due to holes,Ie= current due to electrons, I= total current)
During formation of p-n junction
an electron diffuses from n to p side
a hole diffuses from p to n side
a hole diffuses from n to p side
an electron diffuses from p to n side
What is the name of this component?
PN junction
LED
silicon
PNP
Which of the following statements is incorrect for the depletion region of a diode?
There the mobile charges exist.
Equal number of holes and electrons exist, making the region neutral.
Recombination of holes and electrons has taken place.
None of these
Potential barrier developed in a junction diode opposes the flow of
minority carrier in both regions only
majority carriers only
majority carriers only
holes in p region
If the energy of a photon of sodium light (A = 589 nm) equals the band gap of semiconductor, the minimum energy required to create hole electron pair
1.1 eV
2.1 eV
3.2 eV
1.5 eV
The mass of nitrogen-14 is 14.003074u .
What is this in kilograms?
2.0×10−26kg
2.1×10−26kg
2.2×10−26kg
2.3×10−26kg
What is the mass defect for N714 ?
mp=1.007276u
mn=1.008664u
mN−14=14.003074u
0.10850u
0.10851u
0.10852u
0.10853u
Calculate the energy released by a mass defect of 0.1085u
1.3×10−11J
1.4×10−11J
1.5×10−11J
1.6×10−11J
Calculate the binding energy per nucleon for O810 with mass 17.99916u .
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7.3MeV
7.4MeV
7.5MeV
7.6MeV
