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Physics NEET MCQ
Quiz 10
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Q.1
The wavelenght λ of the Kα X-ray line of an anticathode element of atomic number z is nearly proportional to ...[ MPPMT 1987]
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a)Z2
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b) (Z - 1)2
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c)1/Z-1
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d)1 / (Z-1)2
Explanation
Answer: (d)
Q.2
Both γ-rays and X-rays are electromagnetic waves. the basic difference in them is [ raj.PMT 1997]
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a) The production of X-rays is a nuclear property whereas the production of γ-rays is an atomic property
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b) The production of X-rays is an atomic property whereas the production of γ-rays is a nuclear property
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c)γ rays have less frequency than X-rays
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d)the velocity of γ rays is more than that of X-rays
Explanation
Answer: (b)
Q.3
X-rays passing through a strong uniform magnetic field [ CPMT 1998]
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a) get deflected in the direction of the field
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b) get deflected in the direction opposite to that of field
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c)get deflected in the direction perpendicular to that of the field
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d)do not get deflected at all
Explanation
Answer:(d)
Q.4
The X-ray beam coming from an X-ray tube will be [ IIT 1985]
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a) Monochromatic
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b) Haing all wavelengths smaller than a certain maximum wavelength
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c) Having all wavelengths larger than a certain minimum wavelength
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d) HAving all wavelengths lying between a minimum and maximum wavelength
Explanation
Answer: (c)
Q.5
What happens when fast moving electrons are stopped and fall on the metallic target in an evacuated glass bulb [ MPPMT 1997]
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a)β particles are produced
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b) Metal become soft
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c)γ rays are produced
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d)X-rays are produced
Explanation
Answer: (d)
Q.6
X ray is ...[ CPMT 1999]
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a) Phenomenon of conservation of Kinetic energy into radiation energy
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b) Conservation of momentum
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c)Conservation of mass into energy
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d)principal of conservation of charge
Explanation
Answer: (a)
Q.7
Bragg's equation will not have solution if ....
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a) λ > 2d
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b)λ < 2d
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c)λ < d
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d)λd
Explanation
Answer:(a)
Q.8
Kα characteristic X-ray refers to the transition ..[ MPPMT 1999]
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a) n=2 to n=1
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b) n=3 to n=2
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c) n=3 to n=1
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d) n=4 to n=2
Explanation
Answer: (a)
Q.9
The potential difference applied to an X-ray tube is increased. As a result, in the emitted radiation [ ISM Dhanbad]
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a)the intensity increases
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b) the minimum wavelength increases
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c)the maximum wavelength increases
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d)the minimum wavelength decreases but intensity remains same
Explanation
Answer: (d)
Q.10
When a beam of accelerated electrons hits a target which of the following wavelengths is absent in the X-ray region of spectrum, if the X-ray tube is operating at 40000 volt [ MNR 1995]
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a) 1.5 Å
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b) 0.5 Å
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c)0.25 Å
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d)1.0 Å
Explanation
from the formula λmin=ch/eVHEnce wavelength lower than 0.3Å i.e. 0.25Å will be absent from the spectrumAnswer: (c)
Q.11
The ratio of the energy of an X-ray photon of wavelength 1 Å to that of visible light of wavelength 5000 Å is [ EAMCET 1995]
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a) 1 : 5000
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b) 5000 : 1
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c) 1 : 25 ×106
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d) 25 × 106
Explanation
Answer: (b)
Q.12
The frequency of Kα line for an element of Z=64 is Vα and that for an element Z=80 is v'α. The ratio of frequencies of Kα lines of these elements is
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a) √(2/5)
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b) 2/√5
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c)16/25
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d)2/5
Explanation
From the formula Answer: (c)
Q.13
The wavelength of Kα lines given by molybdenum ( atomic number=42) is 0.7078Å then wavelength of Kα for zinc ( atomic number 30) will b [ raj. PET 1997]
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a) 0.354 Å
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b) 1.4147 Å
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c)09425 Å
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d)1.2547 Å
Explanation
from Moseley's law we have (Z-1)2=ν ∝ 1/λAnswer: (b)
Q.14
Velocity of photon is proportional to [ CBSE 1996] ν is frequency
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a) ν
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b) ν1/2
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c)ν-1/2
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d) ν0
Explanation
Answer:(d)
Q.15
For photoelectric emission from certain metal the cut-off frequency is ν. If radiation of frequency 2ν impinges on the metal plate, the maximum possible velocity of the emitted electron will be (m is the electron mass)…[NEET 2013]
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a)
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b)
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c)
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d)
Explanation
From equation Answer:(c)
Q.16
The wavelength λe of an electron and λp of aphoton of same energy E are related by
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a)
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b)
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c)
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d)
Explanation
We will use de broglie formula for proton On squaring (i) and substituting value of E from (ii) Answer:(a)
Q.17
The wavelength associated with a gold ball weighing 200g and moving at speed of 5m/h is of the order of ( h=6.625 × 10⁻³⁴ Js) [ IIT 2001]
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a) 10-10 m
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b) 10-20 m
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c) 10-30 m
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d) 10-40 m
Explanation
use formula λ=h/mvAnswer: (c)
Q.18
The momentum of pfoton of an electromagnetic radiation is 3.3×10⁻²⁹ kgms-What is the frequaency of the associated waves? [ CBSE-PMT 1990] h=6.6×10⁻³⁴Js; c=3×108 ms-1
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a)1.5 × 1013 Hz
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b) 7.5 × 1012 Hz
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c)6.0×103
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d)3.0×103
Explanation
As λ=h/p and λ=c/ν soν=cP/hby substituting the values we gettν=1.5× 1013 HzAnswer: (a)
Q.19
Monochromatic light of frequaency 6.0×1014 Hz is produced by a laser. the power emitted is 2×10⁻³ W. The number of photons emitted, on the average, by the sources per second is ... [ CBSE-PMT 2007]
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a) 5×1016
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b) 5×1017
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c) 5×1014
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d) 5×1015
Explanation
Since power p=nhν here n is number of photons per second ∴ n=p/ hν Answer: (d)
Q.20
A particle of mass 1mg has the same wave length as an electron moving with velocity of 3×106 ms-The velocity of the particle is [ CBSE-PMT 2008] mass of electron=9.1×10⁻³¹ kg
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a)2.7×10⁻¹⁸ m/s
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b) 9×10⁻² m/s
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c)3×10⁻³¹m/s
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d)2.7×10⁻²¹m/s
Explanation
let m1 be the mass of particle, and velocity be v1me be mass of electron , and ve be the velocity of electronSince wave length is same thus h/p1=h/ pe ∴ m1 v1=me ve v1=me ve / m1on substitution we get Answer:(d)
Q.21
Given that a photon of light of wavelength 10,000Å has an energy equal to 1.23 eV, when light of wave length 5000Å and intensity I0 falls on photoelectric cell and the saturation current is 0.40×10⁻⁶ ampere stopping potential is 1.36 V. Then the work function is .. [ CPMT 1977]
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a) 0.43 eV
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b) 1.10eV
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c)1.36eV
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d)2.47eV
Explanation
Given if wave length of light 10000Å then energy=1.23 eV∴Incident energy of 5000Å=1.23×2=2.46 Incident energy=Work function + Kinetic energy (maximum) hν=Φ + eV0 Φ=hν - eV0 Φ=(2.46 - 1.36)eV=1.10eV Answer:(b)
Q.22
Given Plank's constant h=6.6×10⁻³⁴ J-sec. The momentum of each photon in a given radiation is 3.3 ×10⁻²⁹ kg-m/sec. The wave-length of radiation is
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a) 3 × 10⁻³ m
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b) 6 × 10⁻¹⁰ m
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c)7.5 × 10⁻² m
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d)2 × 10⁻⁵ m
Explanation
Use formula λ=h/pAnswer: (d)
Q.23
The de-Broglie wavelength associated with moving electrons is 0.24×10⁻¹⁰ m. The voltage applied between grids to bring it to rest is
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a) 1000 volts
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b) 2597 volts
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c)2597 volts
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d)none of above
Explanation
Energy of grids must be equal to kinetic energy of electronAnswer: (b)
Q.24
When light of intensity 1 W/m2 and wavelength 5×10⁻⁷ is incident on a surface. It is completely absorbed by the surface. If 100 photons emit one electron and area of surface is 1cm2, then the photoelectric current will be
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a) 2mA
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b) 0.4 µA
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c) 4.0 mA
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d) 4 µA
Explanation
Energy incident on 1sqcm surface=10-4 J IF n are the number of photons striking 1sqcm area then energy E=nhc/λ Only one percent of incident photons emits electrons thus number of electrons emitted N=0.2525 ×1013 Now Current I=Ne/t here t=1 sec I=0.2525×1013=0.4 µA Answer: (b)
Q.25
If 5% of the energy supplied to a bulb is radiated as visible light, how many quanta are emitted per second by a 100 watt lamp? Assume wavelength of visible light as 5.6×10⁻⁵ cm
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a) 1.4 × 1019
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b) 2.0 × 10⁻⁴
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c) 1.4 × 10⁻¹⁹
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d) 2.0 × 104
Explanation
energy of radiation E=5 J use equation E=nhc/λ Answer: (a)
Q.26
When α-particles are accelerated under the p.d of V volt, their de-Broglie's wavelength is ...Å [ Mass of alpha particle=6.4 ×10⁻²⁷ kg and its charge is 3.2×10⁻¹⁹ C.
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a)0.287/√V
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b) 12.27/√V
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c)0.103/√V
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d)1.22/√V
Explanation
use formula λ=h/ √(2mqV) Answer: (c)
Q.27
To reduce de_broglie wavelength of an electron from 10-10m to 0.5×10⁻¹⁰m, its energy should be ...
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a) increased to four times
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b) doubled
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c)halved
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d)decreased
Explanation
Use formula λ=h/ √(2mK) λ ∝ 1/√K Answer: (a)
Q.28
The work function of a metallic substance is 4.0 eV. The longest wavelength of light that can cause photoelectron emission from the substance is approximately.. [ AFMC 1998]
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a)220 nm
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b) 310 nm
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c)400 nm
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d)540 nm
Explanation
Work function Φ=hc/λλ=hc/Φgiven Φ=4 × 10⁻¹⁶ J Answer:(b)
Q.29
The wave length of a particle having a momentum of 2×10⁻²⁸ kgm/s is ..[ AFMC 2002]
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b) 3.3×105 m
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d) 30 m
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a) 3.3×10⁻⁶m
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c) 3.3×10⁻⁴m
Explanation
Wave length λ = h/p λ = 6.6×10⁻³⁴ / 2×10⁻⁶ λ = 3.3×10⁻⁶ Answer: (a)
Q.30
What is the de-Broglie wavelength of 1 kg mass moving with a velocity of 10 m/s? [ AFMC 2001]
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d) none of these
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a) 6.626×10⁻³⁵ m
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b) 6.626×10⁻³³ m
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c) 6.626×10⁻³⁴ m
Explanation
λ = h/p = h/(mv) λ = 6.6×10⁻³⁴ / (1×10) λ = 6.6×10⁻³⁵ Answer:(a)
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