CBSE Questions for Class 12 Medical Physics Dual Nature Of Radiation And Matter Quiz 5 - MCQExams.com

The energy $$E$$ and the momentum $$p$$ of a photon is given by $$E = hv$$ and $$p = \displaystyle\ \frac{h}{\lambda}$$. The velocity of photon will be 
  • $$E/p$$
  • $$Ep$$
  • $$(E/P)^{2}$$
  • $$\sqrt{E/P}$$
$$10^{-3}W$$ of $$5000\overset {o}{A}$$ light is directed on a photoelectric cell. If the current in the cell is $$0.16\mu A$$, the percentage of incident photons which produce photoelectrons, is
  • 40%
  • 0.04%
  • 20%
  • 10%
If 5% of the energy supplied to a bulb is irradiated as visible light, how many quanta are emitted per second by a 100 W lamp? Assume wavelength of visible light of $$5.6\times 10^{-5} cm$$.
  • $$1.4\times 10^{19}$$
  • $$3\times 10^{3}$$
  • $$1.4\times 10^{-19}$$
  • $$3\times 10^{4}$$
The radius of the second orbit of an electron in hydrogen atom is $$2.116\overset {o}{A}$$. The de Broglie wavelength associated with this electron in this orbit would be
  • $$6.64\overset {o}{A}$$
  • $$1.058\overset {o}{A}$$
  • $$2.116\overset {o}{A}$$
  • $$13.28\overset {o}{A}$$
If $$\lambda_1$$ and $$\lambda_2$$ denote the wavelengths of de Broglie waves for electrons in the first and second Bohr orbits in a hydrogen atom, then $$\lambda_1/\lambda_2$$ is equal to :
  • $$2$$
  • $$\dfrac{1}{2}$$
  • $$\dfrac{1}{4}$$
  • $$4$$
A particle of mass M at rest decays into two masses $$m_1$$ and $$m_2$$ with non-zero velocities. The ratio $$\lambda_1 / \lambda_2$$ of de Broglie wavelengths of particles is
  • $$m_2/m_1$$
  • $$m_1/m_2$$
  • $$\sqrt {m_1}/\sqrt {m_2}$$
  • 1:1
A particle of mass m is projected from ground with velocity u making angle $$\theta$$ with the vertical. The de Broglie wavelength of the particle at the highest point is
  • $$\infty$$
  • $$h/mu sin\theta$$
  • $$h/mu cos\theta$$
  • $$h/mu$$
What is the wavelength of a photon of energy 1eV?
  • $$12.4\times 10^3\overset {o}{A}$$
  • $$2.4\times 10^3\overset {o}{A}$$
  • $$0.4\times 10^2\overset {o}{A}$$
  • $$1000\overset {o}{A}$$
How many photons of a radiation of wavelength $$\lambda=5\times 10^{-7}$$ m must fall per second on a blackened plate in order to produce a force of $$6.62\times 10^{-5}N$$?
  • $$3\times 10^{19}$$
  • $$5\times 10^{22}$$
  • $$2\times 10^{22}$$
  • $$1.67\times 10^{18}$$
A particle of mass 3m at rest decays into two particles of masses m and 2m having non-zero velocities. The ratio of the de Broglie wavelengths of the particles $$(\lambda_1/\lambda_2)$$ is
  • 1/2
  • 1/4
  • 2
  • None of these
How many photons are emitted per second by a 5 mW laser source operating at 632.8 nm?
  • $$1.6\times 10^{16}$$
  • $$1.6\times 10^{13}$$
  • $$1.6\times 10^{10}$$
  • $$1.6\times 10^{3}$$
The de Broglie wavelength of a thermal neutron at $$927^oC$$ is $$\lambda$$. Its wavelength at $$327^oC$$ will be
  • $$\lambda /2$$
  • $$\lambda /\sqrt 2$$
  • $$\lambda \sqrt 2$$
  • $$2\lambda$$
The energy of a photon is equal to the kinetic energy of a proton. The energy of photon is $$E$$. Let $$\lambda_1$$ be the de Broglie wavelength of the proton and $$\lambda_2$$ be the wavelength of the photon. Then, $$\lambda_1/\lambda_2$$ is proportional to :
  • $$E^0$$
  • $$E^{1/2}$$
  • $$E^{-1}$$
  • $$E^{-2}$$
A helium-neon laser has a power output of 1 mW of light of wavelength 632.8 nm. Calculate the energy of each photon in eV.
  • 2.5
  • 1.96
  • 0.53
  • 3.3
Two hydrogen atoms are in excited state with electrons residing in $$n=2$$. The first one is moving toward left and emits a photon of energy $$E_1$$ toward right. The second one is moving toward right with the same speed and emits a photon of energy $$E_2$$ toward left. Taking recoil of nucleus into account, during the emission process
  • $$E_1>E_2$$
  • $$E_1< E_2$$
  • $$E_1=E_2$$
  • Information insufficient
A 100 W point source emits monochromatic light of wavelength $$6000\overset {o}{A}$$. Calculate the total number of photons emitted by the source per second.
  • $$5\times 10^{20}$$
  • $$8\times 10^{20}$$
  • $$6\times 10^{20}$$
  • $$3\times 10^{20}$$
An electron and a photon have same wavelength. If $$p$$ is the momentum of electron and $$E$$ is the energy of photon, the magnitude of $$p/E$$ in $$S\space I$$ unit is 
  • $$3.0\times10^8$$
  • $$3.33\times10^{-9}$$
  • $$9.1\times10^{-31}$$
  • $$6.64\times10^{-34}$$
An electron is in an excited state in a hydrogen like atom. It has a total energy of $$-3.4\space eV$$. The kinetic energy of electron is $$'E\ '$$ and its de Broglie wavelength is $$'\lambda\ '$$
  • $$E = 6.8\space eV;\quad \lambda = 6.6\times10^{-10}\space m$$
  • $$E = 3.4\space eV;\quad \lambda = 6.6\times10^{-10}\space m$$
  • $$E = 3.4\space eV;\quad \lambda = 6.6\times10^{-11}\space m$$
  • $$E = 6.8\space eV;\quad \lambda = 6.6\times10^{-11}\space m$$
The energy of photon of green by Potential DIfference of $$5000A^0$$. is
  • $$3.459 \times 10^{-19} joule $$
  • $$3.973 \times 10^{-19} joule $$
  • $$4.132 \times 10^{-19} joule $$
  • $$8453 \times 10^{-19} joule $$
An electron of mass 'm' and charge 'w' initially at rest gets accelerated by a constant electric field 'E'. The rate of change of de-Broglie wavelength of this electron at time 't' ignoring relativistic effects is
  • $$-\dfrac{h}{eEt^2}$$
  • $$-\dfrac{eht}{E}$$
  • $$-\dfrac{mh}{eEt^2}$$
  • $$\dfrac{h}{eE}$$
We wish to see inside an atom. Assume the atom to have a diameter of 100 pm. This means that one must be able to resolve a width of say 10 pm. If an electron microscope is used the energy required should be
  • 1.5 keV
  • 15 keV
  • 150 keV
  • 1.5 MeV
The energy of photon of wavelength $$\lambda $$ is
  • $$c \lambda /h$$
  • $$h \lambda /c$$
  • $$hc/ \lambda$$
  • $$ \lambda/ hc $$
The wavelength of a wave is $$\lambda$$ = 6000 $$\overset{o}{A}$$, then wave number will be
  • 1.66 $$\times$$ 10$$^{7}$$ m$$^{-1}$$
  • 1.66 $$\times$$ 10$$^{6}$$ m$$^{-1}$$
  • 16.6 $$\times$$ 10$$^{-1}$$ m$$^{-1}$$
  • 166 $$\times$$ 10$$^{3}$$ m$$^{-1}$$
A proton and $$\alpha$$-particle are accelerated through the same potential difference. The ratio of their  de-Broglie wavelength will be
  • 1 : 1
  • 1 : 2
  • 2 :1
  • $$2\sqrt 2 : 1$$
If the kinetic energy of a moving particle is E, then the de-Broglie wavelength is
  • $$\lambda =h \sqrt{2 m E}$$
  • $$\lambda =\sqrt{\dfrac{2 m E}{h}}$$
  • $$\lambda =\dfrac{h}{\sqrt{2 m E}}$$
  • $$\lambda =\dfrac{hE}{\sqrt{2 m E}}$$
If a photon and an electron have same de-broglie wavelength, then
  • Both have same kinetic energy
  • Proton has more K.E. than electron
  • Electron has more K.E. than proton
  • Both have same velocity
If the energy of a photon is 10 eV, then its momentum is :
  • $$5.33 \times 10^{-23} kg \,m/s $$
  • $$5.33 \times 10^{-25} kg \,m/s $$
  • $$5.33 \times 10^{-29} kg \,m/s $$
  • $$5.33 \times 10^{-27} kg \,m/s $$
It is essential to consider light as a stream of photons to explain
  • Diffraction of light
  • Refraction of light
  • Photoelectric effect
  • Reflection of light
A 200W sodium street lamp emits yellow light of wavelength $$0.6\mu m$$. Assuming it to be 25% efficient converting electrical energy to light, the number of photons of yellow light it emits per second is :
  • $$62\times 10^{20}$$
  • $$3\times 10^{19}$$
  • $$1.5\times 10^{20}$$
  • $$6\times 10^{18}$$
Two hydrogen atoms are in excited state with electrons residing in $$\displaystyle n=2$$. First one is moving towards left and emits a photon of energy $$\displaystyle { E }_{ 1 }$$ towards right. Second one is moving towards left with same speed and emits a photon of energy $$\displaystyle { E }_{ 2 }$$ towards left. Taking recoil of nucleus into account during emission process, which of the following option is correct?
  • $$\displaystyle { E }_{ 1 }>{ E }_{ 2 }$$
  • $$\displaystyle { E }_{ 1 }<{ E }_{ 2 }$$
  • $$\displaystyle { E }_{ 1 }={ E }_{ 2 }$$
  • information insufficient
In photoelectric effect, ______ present in solar energy changes into electric energy.
  • Only radiant heat
  • Visible light
  • Both radiant heat and light
  • Neither radiant heat nor light
The condition for achieving laser action are
(i) the system must be in a state of population inversion
(ii) the excited state of the system should be in metastable state
(iii) the atom should be in lower energy state
(iv) no conditions required
  • (i) and (ii)
  • (ii) and (iii)
  • (iii) and (iv)
  • (i), (ii), (iii),(iv)
LASER action is found in _________ semiconductor.
  • direct bond gap
  • indirect bond gap
  • germanium
  • silicon
In LASER during the stimulated emission process the photon is __________.
  • lost
  • created
  • absorbed
  • scattered
If the energy of photons corresponding the wavelength of $$6000\mathring { A } $$ is $$3.2\times {10}^{-19}J$$, the photon energy for a wavelength of $$4000\mathring { A } $$ will be
  • $$1.11\times {10}^{-19}J$$
  • $$2.22\times {10}^{-19}J$$
  • $$4.44\times {10}^{-19}J$$
  • $$4.80\times {10}^{-19}J$$
Consider two particles of different masses. In which of the following situations the heavier of the two particles will have smaller de-Broglie wavelength?
  • Both have a free fall through the same height
  • Both moves with the same kinetic energy
  • Both moves with the same linear momentum
  • Both move with the same speed
The momentum of a photon of energy $$1MeV$$ in kg-m/s, will be equal to :
  • $$0.33\times {10}^{6}$$
  • $$7\times {10}^{-24}$$
  • $${10}^{-22}$$
  • $$5\times {10}^{-22}$$
For the Bohr's first orbit of circumference $$2\pi r$$, the de-Broglie wavelength of revolcing electron will be.
  • $$2\pi r$$
  • $$\pi r$$
  • $$\displaystyle\frac{1}{2\pi r}$$
  • $$\displaystyle\frac{1}{4\pi r}$$
The energy of gamma $$\left( \gamma  \right) $$ ray photon is $${ E }_{ \gamma  }$$ and that of an X-ray photon is $${ E }_{ X }$$. If the visible light photon has an energy of $${ E }_{ v }$$, then we can say that:
  • $${ E }_{ X }>{ E }_{ \gamma }>{ E }_{ v }$$
  • $${ E }_{ \gamma }>{ E }_{ v }>{ E }_{ X }$$
  • $${ E }_{ \gamma }>{ E }_{ X }>{ E }_{ v }$$
  • $${ E }_{ X }>{ E }_{ v }>{ E }_{ \gamma }$$
The de-Broglie wavelength of an electron in the ground state of the hydrogen atom is :
  • $$\pi r^2$$
  • $$2\pi r$$
  • $$\pi r$$
  • $$\sqrt{\pi r}$$
In photoelectric effect if the intensity of light is doubled, then maximum kinetic energy of photoelectrons will become.
  • Double
  • Half
  • Four times
  • No change
When the kinetic energy of an electron is increased the wavelength of the associated wave will :
  • Increase
  • Decrease
  • Wavelength does not upon kinetic energy
  • None of the above
The energy of an electron of mass m moving with velocity V and de-Broglie wavelength $$\lambda$$ is __________. ('h' is Planck's constant)
  • $$\displaystyle\frac{h}{2m\lambda}$$
  • $$\displaystyle\frac{h^2}{2m\lambda^2}$$
  • $$\displaystyle\frac{h\lambda}{2m}$$
  • $$\displaystyle\frac{h}{m\lambda}$$
The de-Broglie wavelength of an electron moving with a velocity $$\dfrac{c}{2}$$ (where c is velocity of light in vacuum) is equal to the wavelength of a photon. The ratio of the kinetic energies of electron and photon is:
  • $$1 : 4$$
  • $$1 : 2$$
  • $$1 : 1$$
  • $$2 : 1$$
$$1$$ mole of photon, each of frequency $$2500{ s }^{ -1 }$$, would have approximately a total energy of :
  • $$10 erg$$
  • $$1 J$$
  • $$1 eV$$
  • $$1 MeV$$
Which of the following expression gives the de-Broglie relationship?
  • $$\displaystyle p=\frac { h }{ mv } $$
  • $$\displaystyle \lambda =\frac { h }{ mv } $$
  • $$\displaystyle \lambda =\frac { h }{ mp } $$
  • $$\displaystyle \lambda m=\frac { v }{ p } $$
Hard X-rays for the study of fractures in bones should have a minimum wavelength of $$\displaystyle { 10 }^{ -11 }m$$. The accelerating voltage for electrons in X-ray machine should be:
  • $$<124  \ kV$$
  • $$>124 \ kV$$
  • Between $$60 \ kV $$ and $$ 70\  kV$$
  • $$=100\ kV$$
The energy of a photons is equal to the kinetic energy of a proton. If $$\lambda_1$$ is the de-Broglie wavelength of a proton, $$\lambda_2$$ the wavelength associated with the photon, and if the energy of the photon is E, then $$(\lambda_1/\lambda_2)$$ is proportional to:
  • $$E^4$$
  • $$E^{1/2}$$
  • $$E^2$$
  • $$E$$
The energy that should be added to an electron to reduce its de-Broglie wavelength from $$1$$nm to $$0.5$$nm is.
  • Four times the initial energy
  • Equal to the initial energy
  • Twice the initial energy
  • Thrice the initial energy
Calculate the approximates energy of a photon given that wavelength of photon is 2 nm and Planck constant $$h$$ is $$6.6\times 10^{-34} Js$$.
  • $$4\times 10^{-51}J$$
  • $$1\times 10^{-34}J$$
  • $$1\times 10^{-16}J$$
  • $$1\times 10^{34}J$$
  • $$2\times 10^{-50}J$$
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